Ex 9.4 Class 11 Maths Ncert Solutions

The given series is \( 1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots {n}^{\text {th}} \) term, \( {a}_{{n}}={n}({n}+ 1)\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k(k+1)\)
\(=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k\)
\(=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\)
\(=\frac{n(n+1)}{2}\left[\frac{2 n+1}{3}+1\right]\)
\(=\frac{n(n+1)}{2}\left(\frac{2 n+4}{3}\right)\)
\(=\frac{n(n+1)(n+2)}{3}\)

The given series is \( 1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\ldots {n}^{\text {th}} \) term
\(={a}_{{n}}={n}({n}+1)({n}+2)\)
\(=\left({n}^{2}+{n}\right)({n}+2)\)
\(={n}^{3}+3 {n}^{2}+2 {n}\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}\)
\(=\sum_{k=1}^{n} k^{3}+3 \sum_{k=1}^{n} k^{2}+2 \sum_{k=1}^{n} k\)
\(=\left[\frac{n(n+1)}{2}\right]^{2}+\frac{3 n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2}\)\(=\left[\frac{n(n+1)}{2}\right]^{2}+\frac{n(n+1)(2 n+1)}{2}+n(n+1)\)
\(=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+2 n+1+2\right]\)
\(=\frac{n(n+1)}{2}\left[\frac{n^{2}+n+4 n+6}{2}\right]\)
\(=\frac{n(n+1)}{4}\left(n^{2}+5 n+6\right)\)
\(=\frac{n(n+1)}{4}\left(n^{2}+2 n+3 n+6\right)\)
\(=\frac{n(n+1)[n(n+2)+3(n+2)]}{4}\)
\(=\frac{n(n+1)(n-2)(n-3)}{4}\)

The given series is \( 3 \times 1^{2}+5 \times 2^{2}+7 \times 3^{2}+\ldots {n}^{\text {th}} \) term
\({a}_{{n}}=(2 {n}+1) {n}^{2}=2 {n}^{3}+{n}^{2}\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}\)
\(=\sum_{k=1}^{n}\left(2 k^{3}+k^{2}\right)=2 \sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} k^{2}\)
\(=2\left[\frac{n(n+1)}{2}\right]^{2}+\frac{n(n+1)(2 n+1)}{6}\)
\(=\frac{n^{2}(n+1)^{2}}{2}+\frac{n(n+1)(2 n+1)}{6}\)
\(=\frac{n(n+1)}{2}\left[n(n+1)+\frac{2 n+1}{3}\right]\)
\(=\frac{n(n+1)}{2}\left[\frac{3 n^{2}+3 n+2 n+1}{3}\right]\)
\(=\frac{n(n+1)}{2}\left[\frac{3 n^{2}+5 n+1}{3}\right]\)
\(=\frac{n(n+1)\left(3 n^{2}+5 n+1\right)}{6}\)

The given series is \( \frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots \)
\( {n}^{\text {th}} \) term, an \( =\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\quad \) [By partial fractions]
\( a_{1}=\frac{1}{1}-\frac{1}{2} \)
\( {a}_{2}=\frac{1}{2}-\frac{1}{3} \)
\( {a}_{3}=\frac{1}{3}-\frac{1}{4} \ldots \)
\( {a}_{{n}}=\frac{1}{n}-\frac{1}{n+1} \)
Adding the above terms column wise, we obtain
\(={a}_{1}+{a}_{2}+\ldots+{a}_{{n}}=\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{n}\right]-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots \frac{1}{n+1}\right]\)
\({S}_{{n}}=1-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

The given series is \( 5^{2}+6^{2}+7^{2}+\ldots+20^{2} n^{th}\) term
\({a}_{{n}}=({n}+4)^{2}={n}^{2}+8 {n}+16\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(k^{2}+8 k+16\right)\)
\(=\sum_{k=1}^{n} k^{2}+8 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 16\)
\(=\frac{n(n+1)(2 n+1)}{6}+\frac{8 n(n+1)}{2}+16 n\)
\( 16^{\text {th}} \) term is \( (16+4)^{2}=20^{2} \)
\(S_{n}=\frac{16(16+1)(2 \times 16+1)}{6}+\frac{8 \times 16 \times(16+1)}{2}+16 \times 16\)
\(=\frac{(16)(17)(33)}{6}+\frac{(8)(16)(17)}{2}+16 \times 16\)
\(=1496+1088+256\)
\(=2840\)
\(5^{2}+6^{2}+7^{2}+\ldots+20^{2}=2840\)

The given series is \( 3 \times 8+6 \times 11+9 \times 14+\ldots a^{n }\)
\(=\left({n}^{\text {th}} \text { term of } 3,6,9 \ldots\right) \times\left({n}^{\text {th}}\text { term of } 8,11,14\ldots\right)\)
\(=(3 {n})(3 {n}+5)\)
\(=9 {n}^{2}+15 {n}\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(9 k^{2}+15 k\right)\)
\(=9 \sum_{k=1}^{n} 2+15 \sum_{k=1}^{n} k\)
\(=9 \times \frac{n(n+1)(2 n+1)}{6}+15 \times \frac{n(n+1)}{2}\)
\(=\frac{3 n(n+1)(2 n+1)}{2}+\frac{15 n(n+1)}{2}\)
\(=\frac{3 n(n+1)}{2}(2 {n}+1+5)\)
\(=\frac{3 n(n+1)}{2}(2 n+6)\)
\(=3 {n}({n}+1)({n}+3)\)

The given series is \( 1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{3}\right)+\ldots a^{n }\)
\(=\left(1^{2}+2^{2}+3^{3}+\ldots+n^{2}\right)\)
\(=\frac{n(n+1)(2 n+1)}{6}\)
\(=\frac{n\left(2 n^{2}+3 n+1\right)}{6}=\frac{2^{3}+3 n^{2}+n}{6}\)
\(=\frac{1}{3} n^{3}+\frac{1}{2} n^{2}+\frac{1}{6} n\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}\)
\(=\sum_{k=1}^{n}\left[\frac{1}{3} k^{3}+\frac{1}{2} k^{2}+\frac{1}{6} k\right]\)
\(=\frac{1}{3} \sum_{k=1}^{n} k^{3}+\frac{1}{2} \sum_{k=1}^{n} k^{2}+\frac{1}{6} \sum_{k=1}^{n} k\)
\(=\frac{1}{3} \frac{n^{2}(n+1)^{2}}{(2)^{2}}+\frac{1}{2} \times \frac{n(n+1)(2 n+1)}{6}+\frac{1}{6} \times \frac{n(n+1)}{2}\)
\(=\frac{n(n+1)}{6}\left[\frac{n(n+1)}{2}+\frac{(2 n+1)}{2}+\frac{1}{2}\right]\)
\(=\frac{n(n+1)}{6}\left[\frac{n^{2}+n+2 n+1+1}{2}\right]\)
\(=\frac{n(n+1)}{6}\left[\frac{n^{2}+n+2 n+2}{2}\right]\)
\(=\frac{n(n+1)}{6}\left[\frac{n(n+1)+2(n+1)}{2}\right]\)
\(=\frac{n(n+1)}{6}\left[\frac{(n+1)(n+2)}{2}\right]\)
\(=\frac{n(n+1)^{2}(n+2)}{12}\)

\({a}_{{n}}={n}({n}+1)({n}+4)={n}\left({n}^{2}+5 {n}+4\right)={n}^{3}+5 {n}^{2}+4 {n}\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k\)
\(=\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+\frac{4 n(n+1)}{2}\)
\(=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4\right]\)
\(=\frac{n(n+1)}{2}\left[\frac{3 n^{2}+3 n+20 n+10+24}{6}\right]\)
\(=\frac{n(n+1)}{2}\left[\frac{3 n^{2}+23 n+34}{6}\right]\)
\(=\frac{n(n+1)\left(3 n^{2}+23 n+34\right)}{12}\)

\({a}_{{n}}={n}^{2}+2^{{n}}\)
\({S}_{{n}}=\sum_{k=1}^{n} k^{2}+2^{k}=\sum_{k=1}^{n} k^{2}+\sum_{k-1}^{n} 2^{k} \ldots(1)\)
Consider \( \sum_{k=1}^{n} 2^{k}=2^{1}+2^{2}+2^{3}+\ldots \)
The above series \( 2,2^{2}, 2^{3}, \ldots \) is a G.P. with both the first term and common ratio equal to \(2 \).
\(=\sum_{k=1}^{n} 2^{k}=\frac{(2)\left[(2)^{n}-1\right]}{2-1}=2\left(2^{{n}}-1\right) \ldots(2)\)
Therefore, from (1) and (2), we obtain
\({S}_{{n}}=\sum_{k=1}^{n} k^{2}+2\left(2^{{n}}-1\right)=\frac{n(n+1)(2 n+1)}{6}+2\left(2^{{n}}-1\right)\)

\({a}_{{n}}=(2 {n}-1)^{2}=4 {n}^{2}-4 {n}+1\)
\({S}_{{n}}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(4 k^{2}-4 k+1\right)\)
\(=4 \sum_{k=1}^{n} k^{2}-4 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1\)
\(=\frac{4 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n\)
\(=\frac{2 n(n+1)(2 n+1)}{3}-2 {n}({n}+1)+{n}\)
\(={n}\left[\frac{2\left(2 n^{2}+3 n+1\right)}{3}-2(n+1)+1\right]\)
\(={n}\left[\frac{4 n^{2}+6 n+2-6 n-6+3}{3}\right]\)
\(={n}\left[\frac{4 n^{2}-1}{3}\right]\)
\(=\frac{n(2 n+1)(2 n-1)}{3}\)