Class 11: Special Series – Exercise 21.2

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 1: }3+5+9+15+23+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }3+5+9+15+23+\ldots$
$\displaystyle \text{The successive differences are }2,4,6,8,\ldots$
$\displaystyle \therefore T_n=3+(2+4+6+\ldots+2(n-1))$
$\displaystyle =3+2(1+2+3+\ldots+(n-1))$
$\displaystyle =3+2\left[\frac{(n-1)n}{2}\right]$
$\displaystyle =3+n(n-1)$
$\displaystyle =n^2-n+3$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(k^2-k+3\right)$
$\displaystyle =\sum\limits_{k=1}^{n}k^2-\sum\limits_{k=1}^{n}k+3\sum\limits_{k=1}^{n}1$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+3n$
$\displaystyle =\frac{n(n+1)}{2}\left[\frac{2n+1}{3}-1\right]+3n$
$\displaystyle =\frac{n(n+1)}{2}\left(\frac{2n-2}{3}\right)+3n$
$\displaystyle =\frac{n(n+1)(n-1)}{3}+3n$
$\displaystyle =\frac{n(n^2-1)}{3}+3n$
$\displaystyle =\frac{n^3-n+9n}{3}$
$\displaystyle =\frac{n^3+8n}{3}$
$\displaystyle =\frac{n(n^2+8)}{3}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n^2+8)}{3}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 2: }2+5+10+17+26+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }2+5+10+17+26+\ldots$
$\displaystyle \text{The successive differences are }3,5,7,9,\ldots$
$\displaystyle \therefore T_n=2+(3+5+7+\ldots+(2n-1))$
$\displaystyle =2+\frac{n-1}{2}\left[2(3)+(n-2)(2)\right]$
$\displaystyle =2+\frac{n-1}{2}(2n+2)$
$\displaystyle =2+(n-1)(n+1)$
$\displaystyle =2+n^2-1$
$\displaystyle =n^2+1$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}(k^2+1)$
$\displaystyle =\sum\limits_{k=1}^{n}k^2+\sum\limits_{k=1}^{n}1$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+n$
$\displaystyle =\frac{n(n+1)(2n+1)+6n}{6}$
$\displaystyle =\frac{n\left[(n+1)(2n+1)+6\right]}{6}$
$\displaystyle =\frac{n(2n^2+3n+7)}{6}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(2n^2+3n+7)}{6}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 3: }1+3+7+13+21+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }1+3+7+13+21+\ldots$
$\displaystyle \text{The successive differences are }2,4,6,8,\ldots$
$\displaystyle \therefore T_n=1+(2+4+6+\ldots+2(n-1))$
$\displaystyle =1+2(1+2+3+\ldots+(n-1))$
$\displaystyle =1+2\left[\frac{(n-1)n}{2}\right]$
$\displaystyle =1+n(n-1)$
$\displaystyle =n^2-n+1$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(k^2-k+1\right)$
$\displaystyle =\sum\limits_{k=1}^{n}k^2-\sum\limits_{k=1}^{n}k+\sum\limits_{k=1}^{n}1$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n$
$\displaystyle =\frac{n(n+1)}{2}\left[\frac{2n+1}{3}-1\right]+n$
$\displaystyle =\frac{n(n+1)}{2}\left(\frac{2n-2}{3}\right)+n$
$\displaystyle =\frac{n(n+1)(n-1)}{3}+n$
$\displaystyle =\frac{n(n^2-1)}{3}+n$
$\displaystyle =\frac{n^3-n+3n}{3}$
$\displaystyle =\frac{n^3+2n}{3}$
$\displaystyle =\frac{n(n^2+2)}{3}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n^2+2)}{3}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 4: }3+7+14+24+37+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }3+7+14+24+37+\ldots$
$\displaystyle \text{The successive differences are }4,7,10,13,\ldots$
$\displaystyle \therefore T_n=3+(4+7+10+\ldots)$
$\displaystyle \text{There are }n-1\text{ terms in the A.P. }4,7,10,\ldots$
$\displaystyle \therefore T_n=3+\frac{n-1}{2}\left[2(4)+(n-2)(3)\right]$
$\displaystyle =3+\frac{n-1}{2}(3n+2)$
$\displaystyle =\frac{6+(n-1)(3n+2)}{2}$
$\displaystyle =\frac{6+3n^2-n-2}{2}$
$\displaystyle =\frac{3n^2-n+4}{2}$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(\frac{3k^2-k+4}{2}\right)$
$\displaystyle =\frac{3}{2}\sum\limits_{k=1}^{n}k^2-\frac{1}{2}\sum\limits_{k=1}^{n}k+2\sum\limits_{k=1}^{n}1$
$\displaystyle =\frac{3}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]$
$\displaystyle \qquad-\frac{1}{2}\left[\frac{n(n+1)}{2}\right]+2n$
$\displaystyle =\frac{n(n+1)(2n+1)}{4}-\frac{n(n+1)}{4}+2n$
$\displaystyle =\frac{n(n+1)}{4}\left[(2n+1)-1\right]+2n$
$\displaystyle =\frac{n^2(n+1)}{2}+2n$
$\displaystyle =\frac{n}{2}\left[n(n+1)+4\right]$
$\displaystyle =\frac{n(n^2+n+4)}{2}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n^2+n+4)}{2}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 5: }1+3+6+10+15+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }1+3+6+10+15+\ldots$
$\displaystyle \text{The successive differences are }2,3,4,5,\ldots$
$\displaystyle \therefore T_n=1+(2+3+4+\ldots+n)$
$\displaystyle =1+\frac{n-1}{2}\left[2(2)+(n-2)(1)\right]$
$\displaystyle =1+\frac{n-1}{2}(n+2)$
$\displaystyle =\frac{2+(n-1)(n+2)}{2}$
$\displaystyle =\frac{2+n^2+n-2}{2}$
$\displaystyle =\frac{n^2+n}{2}$
$\displaystyle =\frac{n(n+1)}{2}$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\frac{k(k+1)}{2}$
$\displaystyle =\frac{1}{2}\sum\limits_{k=1}^{n}k^2+\frac{1}{2}\sum\limits_{k=1}^{n}k$
$\displaystyle =\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]$
$\displaystyle \qquad+\frac{1}{2}\left[\frac{n(n+1)}{2}\right]$
$\displaystyle =\frac{n(n+1)}{2}\left[\frac{2n+1}{6}+\frac{1}{2}\right]$
$\displaystyle =\frac{n(n+1)}{2}\left(\frac{2n+4}{6}\right)$
$\displaystyle =\frac{n(n+1)(n+2)}{6}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n+1)(n+2)}{6}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 6: }1+4+13+40+121+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }1+4+13+40+121+\ldots$
$\displaystyle \text{The successive differences are }3,9,27,81,\ldots$
$\displaystyle \therefore T_n=1+\left(3+3^2+3^3+\ldots+3^{n-1}\right)$
$\displaystyle =1+\frac{3(3^{n-1}-1)}{3-1}$
$\displaystyle =1+\frac{3^n-3}{2}$
$\displaystyle =\frac{3^n-1}{2}$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(\frac{3^k-1}{2}\right)$
$\displaystyle =\frac{1}{2}\sum\limits_{k=1}^{n}3^k-\frac{1}{2}\sum\limits_{k=1}^{n}1$
$\displaystyle =\frac{1}{2}\left(3+3^2+3^3+\ldots+3^n\right)-\frac{n}{2}$
$\displaystyle =\frac{1}{2}\left[\frac{3(3^n-1)}{3-1}\right]-\frac{n}{2}$
$\displaystyle =\frac{3^{n+1}-3}{4}-\frac{n}{2}$
$\displaystyle =\frac{3^{n+1}-3-2n}{4}$
$\displaystyle =\frac{3^{n+1}-2n-3}{4}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{3^{n+1}-2n-3}{4}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 7: }4+6+9+13+18+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }4+6+9+13+18+\ldots$
$\displaystyle \text{The successive differences are }2,3,4,5,\ldots$
$\displaystyle \therefore T_n=4+(2+3+4+\ldots+n)$
$\displaystyle =4+\frac{n-1}{2}\left[2(2)+(n-2)(1)\right]$
$\displaystyle =4+\frac{n-1}{2}(n+2)$
$\displaystyle =\frac{8+(n-1)(n+2)}{2}$
$\displaystyle =\frac{8+n^2+n-2}{2}$
$\displaystyle =\frac{n^2+n+6}{2}$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(\frac{k^2+k+6}{2}\right)$
$\displaystyle =\frac{1}{2}\sum\limits_{k=1}^{n}k^2+\frac{1}{2}\sum\limits_{k=1}^{n}k+3\sum\limits_{k=1}^{n}1$
$\displaystyle =\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{1}{2}\left[\frac{n(n+1)}{2}\right]+3n$
$\displaystyle =\frac{n(n+1)}{2}\left[\frac{2n+1}{6}+\frac{1}{2}\right]+3n$
$\displaystyle =\frac{n(n+1)}{2}\left(\frac{2n+4}{6}\right)+3n$
$\displaystyle =\frac{n(n+1)(n+2)}{6}+3n$
$\displaystyle =\frac{n}{6}\left[(n+1)(n+2)+18\right]$
$\displaystyle =\frac{n}{6}\left(n^2+3n+20\right)$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n^2+3n+20)}{6}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 8: }2+4+7+11+16+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }2+4+7+11+16+\ldots$
$\displaystyle \text{The successive differences are }2,3,4,5,\ldots$
$\displaystyle \therefore T_n=2+(2+3+4+\ldots+n)$
$\displaystyle =2+\frac{n-1}{2}\left[2(2)+(n-2)(1)\right]$
$\displaystyle =2+\frac{n-1}{2}(n+2)$
$\displaystyle =\frac{4+(n-1)(n+2)}{2}$
$\displaystyle =\frac{4+n^2+n-2}{2}$
$\displaystyle =\frac{n^2+n+2}{2}$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(\frac{k^2+k+2}{2}\right)$
$\displaystyle =\frac{1}{2}\sum\limits_{k=1}^{n}k^2+\frac{1}{2}\sum\limits_{k=1}^{n}k+\sum\limits_{k=1}^{n}1$
$\displaystyle =\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{1}{2}\left[\frac{n(n+1)}{2}\right]+n$
$\displaystyle =\frac{n(n+1)}{4}\left[\frac{2n+1}{3}+1\right]+n$
$\displaystyle =\frac{n(n+1)}{4}\left(\frac{2n+4}{3}\right)+n$
$\displaystyle =\frac{n(n+1)(n+2)}{6}+n$
$\displaystyle =\frac{n\left[(n+1)(n+2)+6\right]}{6}$
$\displaystyle =\frac{n(n^2+3n+8)}{6}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n^2+3n+8)}{6}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 9: }\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\ldots$
$\displaystyle =\frac{1}{(3\times1-2)(3\times1+1)}+\frac{1}{(3\times2-2)(3\times2+1)}$
$\displaystyle \qquad+\frac{1}{(3\times3-2)(3\times3+1)}+\ldots$
$\displaystyle \therefore T_n=\frac{1}{(3n-2)(3n+1)}$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\frac{1}{(3k-2)(3k+1)}$
$\displaystyle =\frac{1}{3}\sum\limits_{k=1}^{n}\left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)$
$\displaystyle =\frac{1}{3}\left[\left(1+\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3n-2}\right)\right.$
$\displaystyle \qquad\left.-\left(\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3n-2}+\frac{1}{3n+1}\right)\right]$
$\displaystyle =\frac{1}{3}\left(1-\frac{1}{3n+1}\right)$
$\displaystyle =\frac{1}{3}\left(\frac{3n}{3n+1}\right)$
$\displaystyle =\frac{n}{3n+1}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n}{3n+1}.$
$\displaystyle \\$

$\displaystyle \text{Find the sum of the following series to }n\text{ terms:}$
$\displaystyle \textbf{Question 10: }\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+\frac{1}{16\cdot21}+\ldots$
$\displaystyle +\frac{1}{(5n-4)(5n+1)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given series: }\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+\frac{1}{16\cdot21}+\ldots$
$\displaystyle +\frac{1}{(5n-4)(5n+1)}$
$\displaystyle =\frac{1}{(5\times1-4)(5\times1+1)}+\frac{1}{(5\times2-4)(5\times2+1)}$
$\displaystyle \qquad+\frac{1}{(5\times3-4)(5\times3+1)}+\ldots$
$\displaystyle \therefore T_n=\frac{1}{(5n-4)(5n+1)}$
$\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\frac{1}{(5k-4)(5k+1)}$
$\displaystyle =\frac{1}{5}\sum\limits_{k=1}^{n}\left(\frac{1}{5k-4}-\frac{1}{5k+1}\right)$
$\displaystyle =\frac{1}{5}\left[\sum\limits_{k=1}^{n}\frac{1}{5k-4}-\sum\limits_{k=1}^{n}\frac{1}{5k+1}\right]$
$\displaystyle =\frac{1}{5}\left[\left(1+\frac{1}{6}+\frac{1}{11}+\ldots+\frac{1}{5n-4}\right)\right.$
$\displaystyle \qquad\left.-\left(\frac{1}{6}+\frac{1}{11}+\ldots+\frac{1}{5n-4}+\frac{1}{5n+1}\right)\right]$
$\displaystyle =\frac{1}{5}\left(1-\frac{1}{5n+1}\right)$
$\displaystyle =\frac{1}{5}\left(\frac{5n}{5n+1}\right)$
$\displaystyle =\frac{n}{5n+1}$
$\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n}{5n+1}.$
$\displaystyle \\$