Exercise 4.1 Class 11 Maths Solutions

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1+3+3^{2}+\ldots+3^{\mathrm{n}-1}=\frac{\left(3^{n}-1\right)}{2}\)
For \( \mathrm{n}=1 \), we have
\( P(1):=\frac{\left(3^{1}-1\right)}{2}=\frac{3-1}{2}=\frac{2}{2}=1 \), which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer k, i.e.,
\(1+3+3^{2}+\ldots+3^{\mathrm{k}-1}=\frac{\left(3^{k}-1\right)}{2} \ldots\)(i)
We shall now prove that \( P(k+1) \) is true.
Consider
\(1+3+3^{2}+\ldots+3 \mathrm{k}-1+3(\mathrm{k}+1)-1\)
\(=\left(1+3+3^{2}+\ldots+3 \mathrm{k}-1\right)+3 \mathrm{k}\)
\(=\frac{\left(3^{k}-1\right)}{2}+3 k\)
\(=\frac{\left(3^{k}-1\right)+2.3^{k}}{2}[\text { using (1)] }\)
\(=\frac{(1+2) 3^{k}-1}{2}\)
\(=\frac{3.3^{k}-1}{2}\)
\(=\frac{3^{k-1}-1}{2}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N .

Let the given statement be \( \mathrm{P(n)} \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1^{3}+2^{3}+3^{3}+\ldots+\mathrm{n}^{3}=\left(\frac{n(n+1)}{2}\right)^{2}\)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P}(1): 1^{3}=1=\left(\frac{1(1+1)}{2}\right)^{2}=\left(\frac{1.2}{2}\right)^{2}=1^{2}=1 \) which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \), i.e.,
\(1^{3}+2^{3}+3^{3}+\ldots+\mathrm{k}^{3}=\left(\frac{k(k+1)}{2}\right)^{2} \ldots\)(1)
We shall now prove that \( P(k+1) \) is true.
Consider
\(1^{3}+2^{3}+3^{3}+\ldots+k^{3}+(k+1)^{3}\)
\(=\left(1^{3}+2^{3}+3^{3}+\ldots .+k^{3}\right)+(k+1)^{3}\)
\(=\left\{\frac{k(k+1)}{2}\right\}^{2}+(k+1)^{3}[\operatorname{using}(1)]\)
\(=\frac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}\)
\(=\frac{k^{3}(k+1)^{3}+4(k+1)^{3}}{4}\)
\(=\frac{(k+1)^{2}\left\{k^{2}+4(k+1)\right\}}{4}\)
\(=\frac{(k+1)^{2}+\left(k^{2}+4 k+4\right)}{4}\)
\(=\frac{(k+1)^{2}+(k+1)^{2}}{4}\)
\(=\frac{(k+1)^{2}(k+1+1)^{2}}{4}\)
\(=\left\{\frac{(k+1)(k+1+1)}{2}\right\}^{2}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P(n}) \) is true for all natural numbers i.e., N .

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\ldots+\frac{1}{(1+2+3+. .+n)}=\frac{2 n}{(n+1)}\)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P}(1): 1=\frac{2.1}{1+1}=\frac{2}{2}=1 \) which is true.
Let \( \mathrm{P(k)} \) be true for some positive integer \( k \), i.e.,
\(1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\ldots+\frac{1}{(1+2+3+. . k)}=\frac{2 k}{k+1} \ldots\)
We shall now prove that \( \mathrm{P(k+1) }\) is true.
Consider
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+. .+k}+\frac{1}{1+2+3+\cdots+k+(k+1)}\)
\(=\left\{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+k}\right\}+\frac{1}{1+2+3+\cdots+k+(k+1)}\)
\(=\frac{2 k}{k+1}+\frac{1}{1+2+3+. .+k+(k+1)}[\operatorname{using}(1)]\)
\(=\frac{2 k}{k+1}+\frac{1}{\left(\frac{(k+1)(k+1+1)}{2}\right)}\left[1+2+3+. .+n=\frac{n(n+1)}{2}\right]\)
\(=\frac{2 k}{k+1}+\frac{2}{(k+1)(k+2)}\)
\(=\frac{2}{k+1}\left\{k+\frac{1}{k+2}\right\}\)
\(=\frac{2}{k+1}\left\{\frac{k^{2}+2 k+1}{k+2}\right\}\)
\(=\frac{2}{k+1}\left[\frac{(k+1)^{2}}{k+2}\right]\)
\(=\frac{2(k+1)}{k+2}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N .

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1.2.3+2.3.4+\ldots+\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)=\frac{n(n+1)(n+2)(n+3)}{4}\)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P(1)}: 1.2.3=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.2.3 \cdot 4}{4}=6 \) which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer k, i.e.,
\(1.2.3+2.3.4+\ldots+\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)=\frac{k(k+1)(k+2)(k+3)}{4}\)
We shall now prove that \( \mathrm{P(k+1)} \) is true.
Consider
\(1.2 .3+2.3 .4+\ldots+\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)+(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)\)
\(= \{1.2.3+2.3 .4+\ldots+\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)\}+(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)\)
\(= \frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)[\operatorname{using}(1)]\)
\(= (k+1)\left(k+20(k+3)\left(\frac{k}{4}+1\right)\right.\)
\(= \frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
\(= \frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N .

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1.3+2.3^{2}+3.3^{2}+\ldots+\mathrm{n} .3^{\mathrm{n}}=\frac{(2 n-1) 3^{n+1}+3}{4}\)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P(1)}: 1.3=3 \frac{(2 n-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}=\frac{12}{4}=3 \), which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \), i.e.,
\(1.3+2.3^{2}+3.3^{2}+\mathrm{k} 3^{\mathrm{k}}=\frac{(2 k-1) 3^{k+1}+3}{4} \ldots(1)\)
We shall now prove that \( \mathrm{P}(\mathrm{k}+1) \) is true.
Consider
\(1.3+2.3^{2}+3.3^{3}+\ldots+\mathrm{k}.3^{\mathrm{k}}+(\mathrm{k}+1).3 \mathrm{k}+1\)
\(=\left(1.3+2.3^{2}+3.3^{3}+\ldots+\mathrm{k}.3^{\mathrm{k}}\right)+(\mathrm{k}+1).3 \mathrm{k}+1\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true
\(=\frac{(2 k-1) 3^{k+1}+3}{4}+(k+1) 3^{k+1}[\text { using }(1)]\)
\(=\frac{(2 k-1) 3^{k+1}+3+4(k+1) 3^{k+1}}{4}\)
\(=\frac{3^{k+1}\{2 k-1+4(k+1)\}+3}{4}\)
\(=\frac{3^{k+1}(6 k+3)+3}{4}\)
\(=\frac{3^{k+1} \cdot 3(2 k+1)+3}{4}\)
\(=\frac{3^{(k+1)+1}(2 k+1)+3}{4}\)
\(=\frac{(2(k+1)) 3^{(k+1)+1}+3}{4}\)
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1.2+2.3+3.4+\ldots+\mathrm{n}(\mathrm{n}+1)=\left[\frac{n(n+1)(n+2)}{3}\right]\)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P(1)}: 1.2=2=\frac{1(1+1)(1+2)}{3}=\frac{1.2 .3}{3}=2 \) which is true.
Let \( \mathrm{P(k)} \) be true for some positive integer \( k \), i.e.,
We shall now prove that \( \mathrm{P(k+1)} \) is true.
\(\text { Consider } 1.2+2.3+3.4+\ldots+\mathrm{k} .(\mathrm{k}+1)+(\mathrm{k}+1) .(\mathrm{k}+2)\)
\(=[1.2+2.3+3.4+\ldots+\mathrm{k}(\mathrm{k}+1)]+(\mathrm{k}+1)(\mathrm{k}+2)\)
\(=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)[\operatorname{using}(1)]\)
\(=(\mathrm{k}+1)(\mathrm{k}+2)\left(\frac{k}{3}+1\right)\)
\(=\frac{(k+1)(k+2)(k+3)}{3}\)
\(=\frac{(k+1)(k+1+1)(k+1+2)}{3}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1.3+3.5+5.7+\ldots+(2 \mathrm{n}-1)(2 \mathrm{n}+1)=\frac{n\left(4 n^{2}+6 n-1\right)}{3}\)
For \( \mathrm{n}=1 \), we have
\( P(1): 1.3=3=\frac{1\left(4.1^{2}+6.1-1\right)}{3}=\frac{4+6-1}{3}=\frac{9}{3}=3 \), which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer k, i.e.,
\(1.3+3.5+5.7+\ldots+(2 \mathrm{k}-1)(2 \mathrm{k}+1)=\frac{k\left(4 k^{2}+6 k-1\right)}{3} \ldots\)(1)
We shall now prove that \(\mathrm{ P(k+1)} \) is true.
Consider
\((1.3+3.5+5.7+\ldots+(2 \mathrm{k}-1)(2 \mathrm{k}+1)+\{2(\mathrm{k}+1)-1\}\{2(\mathrm{k}+1)+1\}\)
\(=\frac{k\left(4 k^{2}+6 k-1\right)}{3}+(2 k+2-1)(2 k+2+1)[\text { using }(1)]\)
\(=\frac{k\left(4 k^{2}+6 k-1\right)}{3}+(2 k+1)(2 k+3)\)
\(=\frac{k\left(4 k^{2}+6 k-1\right)}{3}+\left(4 k^{2}+8 k+3\right)\)
\(=\frac{k\left(4 k^{2}+6 k-1\right)+3\left(4 k^{2}+8 k+3\right)}{3}\)
\(=\frac{4 k^{3}+6 k^{2}-k+12 k^{2}+24 k+9}{3}\)
\(=\frac{4 k^{3}+18 k^{3}+23 k+9}{3}\)
\(=\frac{4 k^{3}+14 k^{3}+9 k+4 k^{3}+14 k+9}{3}\)
\(=\frac{k\left(4 k^{2}+14 k+9\right)+1\left(4 k^{2}+14 k+9\right)}{3}\)
\(=\frac{(k+1)\left(4 k^{2}+14 k+9\right)}{3}\)
\(=\frac{(k+1)\left\{4 k^{2}+8 k+4+6 k+6-1\right\}}{3}\)
\(=\frac{(k+1)\left\{4\left(k^{2}+2 k+1\right)+6(k+1)-1\right\}}{3}\)
\(=\frac{(k+1)\left\{4(k+1)^{2}+6(k+1)-1\right\}}{3}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N .

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1.2+2.2^{2}+3.2^{2}+\ldots+\mathrm{n} .2^{\mathrm{n}}=(\mathrm{n}-1) 2^{\mathrm{n}+1}+2\)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P}(1): 1.2=2=(1-1) 2^{1+1}+2=0+2=2 \), which is true.
Let \( P(k) \) be true for some positive integer \( k \), i.e.,
\(1.2+2.2^{2}+3.2^{2}+\ldots+\mathrm{k} .2^{\mathrm{k}}=(\mathrm{k}-1) 2^{\mathrm{k}+1}+2 \ldots \text { (i) }\)
We shall now prove that \( \mathrm{P}(\mathrm{k}+1) \) is true.
Consider
\(\left\{1 \cdot 2+2 \cdot 2^{2}+3 \cdot 2^{3}+\cdots+k \cdot 2^{k}\right\}+(k+1) \cdot 2^{k+1}\)
\(=(k+1) 2^{k+1}+2+(k+1) 2^{k+1}\)
\(=2^{k+1}\{(k-1)+(k+1)\}+2\)
\(=2^{k+1} \cdot 2 k+2\)
\(=\mathrm{k} \cdot 2^{(k+1)+1}+2\)
\(=\{(k+1)-1\} 2^{(k+1)+1}+2\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N .

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}\)
For \( \mathrm{n}=1 \), we have
\( P(1): \frac{1}{2}=1-\frac{1}{2^{1}}=\frac{1}{2} \) which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \), i.e.,
\(=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^{k}}=1-\frac{1}{2^{k}} \ldots\)(1)
We shall now prove that \( \mathrm{P}(\mathrm{k}+1) \) is true.
Consider
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}\)
\(=\left(1-\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}[\operatorname{using}(1)]\)
\(=1-\frac{1}{2^{k}}+\frac{1}{2.2^{k}}\)
\(=1-\frac{1}{2^{k}}\left(1-\frac{1}{2}\right)\)
\(=1-\frac{1}{2^{k}}\left(\frac{1}{2}\right)\)
\(=1-\frac{1}{2^{k+1}}\)

Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{(6 n+4)}\)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P (1)}:\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10^{\prime}}\) which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \), i.e.,
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots+\frac{1}{(3 k-1)(3 k+2)}=\frac{k}{6 k+4} \ldots\)(1)
We shall now prove that \( P(k+1) \) is true
Consider
\(=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots+\frac{1}{(3 k-1)(3 k+2)}+\frac{1}{\{3(k+1)-1\}\{3(k+1)+2\}}\)
\(=\frac{k}{6 k+4}+\frac{1}{(3 k+3-1)(3 k+3+2)}[\text { using }(1)]\)
\(=\frac{k}{6 k+4}+\frac{1}{(3 k+2)(3 k+5)}\)
\(=\frac{k}{2(3 k+2)}+\frac{1}{(3 k+2)(3 k+5)}\)
\(=\frac{1}{3 k+2}\left(\frac{k}{2}+\frac{1}{3 k+5}\right)\)
\(=\frac{1}{3 k+2}\left(\frac{k(3 k+5)+2}{2(3 k+5)}\right)\)
\(=\frac{1}{3 k+2}\left(\frac{3 k^{2}+5 k+2}{2(3 k+5)}\right)\)
\(=\frac{1}{3 k+2}\left(\frac{(3 k+2)(k+1)}{2(3 k+5)}\right)\)
\(=\frac{k+1}{2(3 k+5)}\)
\(=\frac{k+1}{6 k+10}\)
\(=\frac{k+1}{6(k+1)+4}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( P(n) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): \frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\cdots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}\)
For \( \mathrm{n}=1 \), we have, which is true.
\( P(1): \frac{1}{1.2.3}=\frac{1 .(1+3)}{4(1+1)(1+2)}=\frac{1.4}{4.2.3}=\frac{1}{1.2 .3^{\prime}} \) which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \), i.e.,
\(\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\cdots+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)} \ldots(1)\)
We shall now prove that \( \mathrm{P}(\mathrm{k}+1) \) is true.
Consider
\(=\left[\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\cdots+\frac{1}{k(k+1)(k+2)}\right]+\frac{1}{(k+1)(k+2)(k+3)}\)
\(=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}[\operatorname{using}(1)]\)
\(=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+3)}{4}+\frac{1}{k+3}\right\}\)
\(=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+3)^{2}+4}{4(k+3)}\right\}\)
\(=\frac{1}{(k+1)(k+2)}\left\{\frac{k\left(k^{2}+6 k+9\right)+4}{4(k+3)}\right\}\)
\(=\frac{1}{(k+1)(k+2)}\left\{\frac{k^{3}+6 k^{2}+9 k+6}{4(k+3)}\right\}\)
\(=\frac{1}{(k+1)(k+2)}\left\{\frac{k^{3}+2 k^{3}+k+4 k^{3}+8 k+4}{4(k+3)}\right\}\)
\(=\frac{1}{(k+1)(k+2)}\left\{\frac{k\left(k^{2}+2 k+1\right)+4\left(k^{2}+2 k+1\right)}{4(k+3)}\right\}\)
\(=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+1)^{2}+4(k+1)^{2}}{4(k+3)}\right\}\)
\(=\frac{(k+1)^{2}(k+4)}{4(k+1)(k+2)(k+3)}\)
\(=\frac{(k+1)\{(k+1)+3\}}{4\{(k+1)+1\}\{(k+1)+2\}}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\( \mathrm{P}(\mathrm{n}): a+a r+a r^{2}+\cdots+a r^{n-1}=\frac{a\left(r^{n}-1\right)}{r-1} \)
For \( \mathrm{n}=1 \), we have
\( \mathrm{P}(1): a=\frac{a\left(r^{1}-1\right)}{r-1}=a \), which is true.
Let \( P(k) \) be true for some positive integer \( k \), i.e.,
\(=a+a r+a r^{2}+\cdots+a r^{k-1}=\frac{a\left(r^{k}-1\right)}{r-1} \ldots \text { (1) }\)
We shall now prove that \( P(k+1) \) is true.
Consider
\(\left\{a+a r+a r^{2}+\cdots+a r^{k-1}\right\}+a r^{(k+1)-1}\)
\(=\frac{a\left(r^{k}-1\right)}{r-1}+a r^{k}[\text { using }(1)]\)
\(=\frac{a\left(r^{k}-1\right)+a r^{k}(r-1)}{r-1}\)
\(=\frac{a\left(r^{k}-1\right)+a r^{k-1}-a r^{k}}{r-1}\)
\(=\frac{a r^{k}-a+a r^{k+1}-a r^{k}}{r-1}\)
\(=\frac{a r^{k+1}-a}{r-1}\)
\(=\frac{a\left(r^{k+1}-1\right)}{r-1}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( P(n) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(P(n):=\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 n+1)}{n^{2}}\right)=(n+1)^{2}\)
For \( \mathrm{n}=1 \), we have
\( P(1):\left(1+\frac{3}{1}\right)+4=(1+1)^{2}=2^{2}=4 \), which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer k, i.e.,
\(=\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 k+1)}{k^{2}}\right)=(k+1)^{2} \ldots\)(1)
We shall now prove that \( P(k+1) \) is true.
Consider
\({\left[\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 k+1)}{k^{2}}\right)\right]\left\{1+\frac{(2(k+1)+1)}{(k+1)^{2}}\right\}}\)
\(=(k+1)^{2}\left(1+\frac{2(k+1)+1}{(k+1)^{2}}\right)[\operatorname{using}(1)]\)
\(=(k+1)^{2}\left[\frac{(k+1)^{2}+2(k+1)+1}{(k+1)^{2}}\right]\)
\(=(\mathrm{k}+1)+2(\mathrm{k}+1)+1\)
\(=\{(\mathrm{k}+1)+1\}+2\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(P(n):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)=(n+1)\)
For \( \mathrm{n}=1 \), we have
\( P(1):\left(1+\frac{1}{1}\right)=2=(1+1) \), which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \), i.e.,
\(\mathrm{P}(\mathrm{k}):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{k}\right)=(k+1) \ldots\)(1)
We shall now prove that \( P(k+1) \) is true.
Consider
\({\left[\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{k}\right)\right]\left(1+\frac{1}{k+1}\right)}\)
\(=(k+1)\left[1+\frac{1}{k+1}\right][\operatorname{using}(1)]\)
\(=(k+1)\left\{\frac{(k+1)+1}{(k+1)}\right\}\)
\(=(k+1)+1\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(P(n): 1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\frac{n(2 n-1)(2 n+1)}{3}\)
For \( \mathrm{n}=1 \) we have
\( \mathrm{P}(1): 1^{2}=1=\frac{1(2.1-1)(2.1+1)}{3}=\frac{1.1 .3}{3}=1 \), which is true
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer k, i.e.,
\(\mathrm{P}(\mathrm{k}): 1^{2}+3^{2}+5^{2}+(2 k+1)^{2}=\frac{k(2 k-1)(2 k+1)}{3} \ldots\)(1)
We shall now prove that \( P(k+1) \) is true.
Consider
\(\left\{1^{2}+3^{2}+5^{2}+\cdots+(2 k-1)^{2}\right\}+\{2(k+1)-1\}^{2}\)
\(=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+2-1)^{2}[\text { using }(1)]\)
\(=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+1)^{2}\)
\(=\frac{k(2 k-1)(2 k+1)+3(2 k+1)^{2}}{3}\)
\(=\frac{(2 k+1)\{k(2 k-1)+3(2 k+1)\}}{3}\)
\(=\frac{(2 k+1)\left\{2 k^{2}-k+6 k+3\right\}}{3}\)
\(=\frac{(2 k+1)\left\{2 k^{2}+5 k+3\right\}}{3}\)
\(=\frac{(2 k+1)\left(2 k^{2}+2 k+3 k+3\right)}{3}\)
\(=\frac{(2 k+1)\{2 k(k+1)+3(k+1)\}}{3}\)
\(=\frac{(2 k+1)(k+1)(2 k+3)}{3}\)
\(=\frac{(k+1)\{2(k+1)-1\}\{2(k+10+1\}}{3}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{(3 n+1)}\)
For \( \mathrm{n}=(1) \) we have
\( \mathrm{P}(1)=\frac{1}{1.4}=\frac{1}{3.1+4}=\frac{1}{4}=\frac{1}{1.4^{\prime}} \) which is true.
Let \( P(k) \) be true for some positive integer \( k \), i.e.,
\(\mathrm{P}(\mathrm{k})=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots+\frac{1}{(3 k-2)(3 k+1)}=\frac{k}{(3 k+1)} \ldots(1)\)
We shall now prove that \( \mathrm{P}(\mathrm{k}+1) \) is true.
Consider
\(=\left\{\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots+\frac{1}{(3 k-2)(3 k+1)}\right\}+\frac{1}{\{3(k+1)-2\}\{3(k+1)+1\}}\)
\(=\frac{k}{3 k+1}+\frac{1}{(3 k+1)(3 k+4)}[\operatorname{using}(1)]\)
\(=\frac{1}{(3 k+1)}\left\{k+\frac{1}{(3 k+4)}\right\}\)
\(=\frac{1}{(3 k+1)}\left\{\frac{k(3 k+4)+1}{(3 k+4)}\right\}\)
\(=\frac{1}{(3 k+1)}\left\{\frac{3 k^{2}+4 k+1}{(3 k+4)}\right\}\)
\(=\frac{1}{3 k+1}\left\{\frac{3 k^{2}+3 k+k+1}{(3 k+4)}\right\}\)
\(=\frac{(3 k+1)(k+1)}{(3 k+1)(3 k+4)}\)
\(=\frac{(k+1)}{3(k+1)+1}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\(\mathrm{P}(\mathrm{n}): \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\cdots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\)
For \( \mathrm{n}=1 \), we have
\(P (1): \frac{1}{3.5}=\frac{1}{3(2.1+3)}=\frac{1}{3.5^{\prime}} \) which is true.
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \), i.e.,
\(\mathrm{P}(\mathrm{k}): \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\cdots+\frac{1}{(2 k+1)(2 k+3)}=\frac{k}{3(2 k+3)} \ldots(1)\)
We shall now prove that \( \mathrm{P}(\mathrm{k}+1) \) is true.
Consider
\(=\left[\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\cdots+\frac{1}{(2 k+1)(2 k+3)}\right]+\frac{1}{\{2(k+1)+1\}\{2(k+1))+3\}}\)
\(=\frac{k}{3(2 k+3)}+\frac{1}{(2 k+3)(2 k+5)}\) [using (1)]
\(=\frac{1}{(2 k+3)}\left[\frac{k}{3}+\frac{1}{(2 k+5)}\right]\)
\(=\frac{1}{(2 k+3)}\left[\frac{k(2 k+5)+3}{3(2 k+5)}\right]\)
\(=\frac{1}{(2 k+3)}\left[\frac{2 k^{2}+5 k+3}{3(2 k+5)}\right]\)
\(=\frac{1}{(2 k+3)}\left[\frac{2 k^{2}+2 k+3 k+3}{3(2 k+5)}\right]\)
\(=\frac{1}{2 k+3}\left[\frac{2 k(k+1)+3(k+1)}{3(2 k+5)}\right]\)
\(=\frac{(k+1)(2 k+3)}{3(2 k+3)(2 k+5)}\)
\(=\frac{(k+1)}{3\{2(k+1)+3\}}\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( P(n) \), i.e.,
\(\mathrm{P}(\mathrm{n}): 1+2+3+\ldots+n < \frac{1}{8}(2 n+1)^{2}\)
It can be noted that \( \mathrm{P}(\mathrm{n}) \) is true for \( \mathrm{n}=1 \) since
\(1 < \frac{1}{8}(2.1+1)^{2}=\frac{9}{8}\)
Let \( P(k) \) be true for some positive integer \( k \), i.e.,
\(\mathrm{P}(\mathrm{k})=1+2+3+\ldots+k < \frac{1}{8}(2 k+1)^{2}\)
We shall now prove that \( P(k+1) \) is true whenever \( P(k) \) is true.
Consider
\((1+2+. .+k)+(k+1) < \frac{1}{8}(2 k+1)^{2}+(k+1)[\text { using }(1)]\)
\(< \frac{1}{8}\left\{(2 k+1)^{2}+8(k+1)\right\}\)
\(< \frac{1}{8}\left\{4 k^{2}+4 k+1+8 k+8\right\}\)
\(< \frac{1}{8}\left\{4 k^{2}+12 k+9\right\}\)
\( < \frac{1}{8}(2 k+3)^{2}\)
\(< \frac{1}{8}\{2(k+1)+1\}^{2}\)
Hence, \( (1+2+3+\cdots+k)+(k+1) < \frac{1}{8}(2 k+1)^{2}+(k+1) \)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\( \mathrm{P}(\mathrm{n}): \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+5) \), which is a multiple of \(3\) .
It can be noted that \( P(n) \) is true for \( \mathrm{n}=1 \) since \( 1(1+1)(1+5)=12 \), which is a multiple of \(3\) .
Hence,
Let \( P(k) \) be true for some positive integer \( k \), i.e.,
\( k(k+1)(k+5) \) is a multiple of \(3\) .
\( \therefore \mathrm{k}(\mathrm{k}+1)(\mathrm{k}+5)=3 \mathrm{~m} \), where \( \mathrm{m} \in \mathrm{N} \ldots\)(1)
We shall now prove that \( P(k+1) \) is true whenever \( P(k) \) is true.
Consider
\((\mathrm{k}+1)\{(\mathrm{k}+1)+1\}\{(\mathrm{k}+1)+5\}\)
\(=(\mathrm{k}+1)(\mathrm{k}+2)\{(\mathrm{k}+5)+1\}\)
\(=(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+5)+(\mathrm{k}+1)(\mathrm{k}+2)\)
\(=\{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+5)+2(\mathrm{k}+1)(\mathrm{k}+5)\}+(\mathrm{k}+1)(\mathrm{k}+2)\)
\(=3 \mathrm{~m}+(\mathrm{k}+1)\{2(\mathrm{k}+5)+(\mathrm{k}+2)\}\)
\(=3 \mathrm{~m}+(\mathrm{k}+1)\{2 \mathrm{k}+10+\mathrm{k}+2\}\)
\(=3 \mathrm{m}+(\mathrm{k}+1)\{3 \mathrm{k}+12\}\)
\(=3 \mathrm{m}+3(\mathrm{k}+1)(\mathrm{k}+4)\)
\( =3 \{\mathrm{m}+(\mathrm{k}+1)(\mathrm{k}+4)\}=3 \times \mathrm{q}\), where \( \mathrm{q}=[\mathrm{m}+(\mathrm{k}+1)(\mathrm{k}+4)\} \) is some natural number.
Therefore, \( (\mathrm{k}+1)\{(\mathrm{k}+1)+1\}\{(\mathrm{k}+1)+5\} \) is a multiple of \(3\).
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.

Let the given statement be \( \mathrm{P}(\mathrm{n}) \), i.e.,
\( \mathrm{P}(\mathrm{n}): 10^{2 \mathrm{n}-1}+1 \) is divisible by 11 .
It can be observed that \( P(n) \) is true for \( \mathrm{n}=1 \)
since \( P(1)=10^{2.1-1}+1=11 \), which is divisible by 11 .
Let \( P(k) \) be true for some positive integer \( k \),
i.e., \( 10^{2 \mathrm{k}-1}+1 \) is divisible by 11 .
\( \therefore 10^{2 \mathrm{k}-1}+1=11 \mathrm{m} \), where \( \mathrm{m} \in \mathrm{N} \ldots\)(1)
We shall now prove that \( P(k+1) \) is true whenever \( P(k) \) is true.
Consider
\(10^{2(\mathrm{k}+1)-1}+1\)
\(=10^{2 \mathrm{k}+2-1}+1\)
\(=10^{2 \mathrm{k}+1}+1\)
\(10^{2}\left(10^{2 \mathrm{k}-1}+1-1\right)+1\)
\(=10^{2}\left(10^{2 \mathrm{k}-1}+1\right)-102+1\)
\(=10^{2} .11 \mathrm{m}-100+1[\text { using }(1)]\)
\(=100 \times 11 \mathrm{m}-99\)
\(=11(100 \mathrm{m}-9)\)
\( =11 \mathrm{r} \), where \( \mathrm{r}=(100 \mathrm{m}-9) \) is some natural number
Therefore, \( 10^{2(k+1)-1}+1 \) is divisible by 11 .
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Hence, by the principle of mathematical induction, statement \( \mathrm{P}(\mathrm{n}) \) is true for all natural numbers i.e., N.