Question

The sum of the series
$\frac{1}{1\times 2}{{}^{25}C}_0+\frac{1}{2\times 3}{{}^{25}C}_1+\frac{1}{3\times 4}{{}^{25}C}_2+...\frac{1}{26\times 27}{{}^{25}C}_{25}$

• A. $\frac{2^{27}-1}{26\times 27}$
• B. $\frac{2^{27}-28}{26\times 27}$
• C. $\frac{1}{2}\left(\frac{2^{26}-1}{26\times 27}\right)$
• D. $\frac{2^{26}-1}{52}$

Answer 1

The correct option is B $\frac{2^{27}-28}{26\times 27}$

Given series:

$\frac{1}{1\times 2}{{}^{25}C}_0+\frac{1}{2\times 3}{{}^{25}C}_1+\frac{1}{3\times 4}{{}^{25}C}_2+...\frac{1}{26\times 27}{{}^{25}C}_{25}$

$\because$ ${\int }_0^x{(1+x)}^{25}dx={\int }_0^x[{{}^{25}C}_0+{{}^{25}C}_{1}x+{{}^{25}C}_2{x}^2+.....+{{}^{25}C}_{25}{x}^{25}]dx$

On integrating w.r. t $x$ taking limits $o$ to $x$ we get
${\left[\frac{{(1+x)}^{26}}{26}\right]}_0^x$

$={[{{}^{25}C}_{0}x+{{}^{25}C}_1\frac{x^2}{2}+{{}^{25}C}_2\frac{x^3}{3}+.....+{{}^{25}C}_{25}\frac{x^{26}}{26}]}_0^x$

$\Rightarrow \{\frac{1}{26}{(1+x)}^{26}-\frac{1}{26}\}$

$={{}^{25}C}_{0}x+{{}^{25}C}_1\frac{x^2}{2}+{{}^{25}C}_2\frac{x^3}{3}+.....+{{}^{25}C}_{25}\frac{x^{26}}{26}$

Again integrating w.r.t $x$ taking limits $0$ to $1$ we get
$\frac{1}{26}{\int }_0^1[{(1+x)}^{26}-1]dx$

$={\int }_0^1[{{}^{25}C}_{0}x+{{}^{25}C}_1\frac{x^2}{2}+{{}^{25}C}_2\frac{x^3}{3}+.....+{{}^{25}C}_{25}\frac{x^{26}}{26}]dx$

$\Rightarrow \frac{1}{26}{[\frac{{(1+x)}^{27}}{27}-x]}_0^1$

$={[{{}^{25}C}_0\frac{x^2}{2}+{{}^{25}C}_1\frac{x^3}{2\times 3}+.....+{{}^{25}C}_{25}\frac{x^{27}}{26\times 27}]}_0^1$

$\Rightarrow \frac{1}{26}\{\frac{2^{27}}{27}-1-\frac{1}{27}\}$

$=\frac{1}{2}{{}^{25}C}_0+\frac{1}{2\times 3}{{}^{25}C}_1+....+\frac{1}{26\times 27}{{}^{25}C}_{25}$

$\therefore \frac{1}{1\times 2}{{}^{25}C}_0+\frac{1}{2\times 3}{{}^{25}C}_1+....+\frac{1}{26\times 27}{{}^{25}C}_{25}=\frac{2^{27}-28}{26\times 27}$