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# Find out the sum of the series$\dfrac{1}{{1 \times 2}}{}^{25}{C_0} + \dfrac{1}{{2 \times 3}}{}^{25}{C_1} + \dfrac{1}{{3 \times 4}}{}^{25}{C_2} + ............ + \dfrac{1}{{26 \times 27}}{}^{25}{C_{25}} \\ {\text{a}}{\text{. }}\dfrac{{{2^{27}} - 1}}{{26 \times 27}} \\ {\text{b}}{\text{. }}\dfrac{{{2^{27}} - 28}}{{26 \times 27}} \\ {\text{c}}{\text{. }}\dfrac{1}{2}\left( {\dfrac{{{2^{26}} - 1}}{{26 \times 27}}} \right) \\ {\text{d}}{\text{. }}\dfrac{{{2^{26}} - 1}}{{52}} \\$  Answer Verified
Hint: - Use${\left( {1 + x} \right)^{25}} = {}^{25}{C_0} + {}^{25}{C_1}x + {}^{25}{C_2}{x^2} + ................. + {}^{25}{C_{25}}{x^{25}}$, then apply integration on both sides with limit 0 to $x$

According to Binomial Theorem the expansion of ${\left( {1 + x} \right)^{25}}$is
${\left( {1 + x} \right)^{25}} = {}^{25}{C_0} + {}^{25}{C_1}x + {}^{25}{C_2}{x^2} + ................. + {}^{25}{C_{25}}{x^{25}}...........\left( 1 \right)$
Integrate equation 1 w.r.t.$x$With limit 0 to $x$
$\int\limits_0^x {{{\left( {1 + x} \right)}^{25}}dx} = \int\limits_0^x {\left( {{}^{25}{C_0} + {}^{25}{C_1}x + {}^{25}{C_2}{x^2} + ................. + {}^{25}{C_{25}}{x^{25}}} \right)dx}$
Let, $\left( {1 + x} \right) = t...................\left( 2 \right)$
So when
$x = 0 \Rightarrow t = 1 + 0 = 1 \\ x = x \Rightarrow t = 1 + x \\$
Now, differentiate equation 2 w.r.t.$x$
$\Rightarrow 0 + dx = dt \Rightarrow dx = dt$
Substitute these value in equation 1
$\int\limits_1^{1 + x} {{t^{25}}dt} = \int\limits_0^x {\left( {{}^{25}{C_0} + {}^{25}{C_1}x + {}^{25}{C_2}{x^2} + ................. + {}^{25}{C_{25}}{x^{25}}} \right)dx}$
Now integrate this equation as you know $\int {{t^n}dt = \left[ {\dfrac{{{t^{n + 1}}}}{{n + 1}}} \right]}$
$\Rightarrow \left[ {\dfrac{{{t^{26}}}}{{26}}} \right]_1^{1 + x} = \left[ {{}^{25}{C_0}x + {}^{25}{C_1}\dfrac{{{x^2}}}{2} + {}^{25}{C_2}\dfrac{{{x^3}}}{3} + ................. + {}^{25}{C_{25}}\dfrac{{{x^{26}}}}{{26}}} \right]_o^x$
Now apply integrating limit
$\left[ {\dfrac{{{{\left( {1 + x} \right)}^{26}}}}{{26}} - \dfrac{1}{{26}}} \right] = \left[ {{}^{25}{C_0}x + {}^{25}{C_1}\dfrac{{{x^2}}}{2} + {}^{25}{C_2}\dfrac{{{x^3}}}{3} + ................. + {}^{25}{C_{25}}\dfrac{{{x^{26}}}}{{26}} - 0 - 0 - 0} \right] \\ \left[ {\dfrac{{{{\left( {1 + x} \right)}^{26}}}}{{26}} - \dfrac{1}{{26}}} \right] = \left[ {{}^{25}{C_0}x + {}^{25}{C_1}\dfrac{{{x^2}}}{2} + {}^{25}{C_2}\dfrac{{{x^3}}}{3} + ................. + {}^{25}{C_{25}}\dfrac{{{x^{26}}}}{{26}}} \right]..............\left( 3 \right) \\$
Now again integrate equation 3 w.r.t.$x$From limit 0 to 1.
$\int\limits_0^1 {\left[ {\dfrac{{{{\left( {1 + x} \right)}^{26}}}}{{26}} - \dfrac{1}{{26}}} \right]dx} = \int\limits_0^1 {\left[ {{}^{25}{C_0}x + {}^{25}{C_1}\dfrac{{{x^2}}}{2} + {}^{25}{C_2}\dfrac{{{x^3}}}{3} + ................. + {}^{25}{C_{25}}\dfrac{{{x^{26}}}}{{26}}} \right]} dx$
Let, $\left( {1 + x} \right) = t...................\left( 4 \right)$
So when
$x = 0 \Rightarrow t = 1 + 0 = 1 \\ x = 1 \Rightarrow t = 1 + 1 = 2 \\$
Now, differentiate equation 4 w.r.t.$x$
$\Rightarrow 0 + dx = dt \Rightarrow dx = dt$
Substitute these value in equation 3
$\int\limits_1^2 {\left[ {\dfrac{{{t^{26}}}}{{26}} - \dfrac{1}{{26}}} \right]dt} = \int\limits_0^1 {\left[ {{}^{25}{C_0}x + {}^{25}{C_1}\dfrac{{{x^2}}}{2} + {}^{25}{C_2}\dfrac{{{x^3}}}{3} + ................. + {}^{25}{C_{25}}\dfrac{{{x^{26}}}}{{26}}} \right]} dx$
Now integrate this equation as you know $\int {{t^n}dt = \left[ {\dfrac{{{t^{n + 1}}}}{{n + 1}}} \right]}$
$\left[ {\dfrac{{{t^{27}}}}{{27 \times 26}} - \dfrac{1}{{26}}} \right]_1^2 = \left[ {{}^{25}{C_0}\dfrac{{{x^2}}}{2} + {}^{25}{C_1}\dfrac{{{x^3}}}{{2 \times 3}} + {}^{25}{C_2}\dfrac{{{x^4}}}{{3 \times 4}} + ................. + {}^{25}{C_{25}}\dfrac{{{x^{27}}}}{{26 \times 27}}} \right]_0^1$
Now apply integrating limit
$\left[ {\dfrac{{{2^{27}}}}{{27 \times 26}} - \dfrac{1}{{26}} - \left( {\dfrac{{{1^{27}}}}{{27 \times 26}}} \right)} \right] = \left[ {{}^{25}{C_0}\dfrac{1}{2} + {}^{25}{C_1}\dfrac{1}{{2 \times 3}} + {}^{25}{C_2}\dfrac{1}{{3 \times 4}} + ................. + {}^{25}{C_{25}}\dfrac{1}{{26 \times 27}} - 0 - 0 - 0} \right] \\ \Rightarrow \left[ {\dfrac{{{2^{27}} - 28}}{{27 \times 26}}} \right] = \dfrac{1}{{1 \times 2}}{}^{25}{C_0} + \dfrac{1}{{2 \times 3}}{}^{25}{C_1} + \dfrac{1}{{3 \times 4}}{}^{25}{C_2} + ............ + \dfrac{1}{{26 \times 27}}{}^{25}{C_{25}} \\$
Hence, option (b) is correct.

Note: - Whenever we face such type of problem the key concept we have to remember is that always remember the Binomial expansion of${\left( {1 + x} \right)^n}$, then integrate the expansion w.r.t.$x$ With limit 0 to x, then again integrate w.r.t.$x$with limit 0 to 1, we will get the required answer.
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Binomial Theorem Class 11  Binomial Theorem  CBSE Class 11 Maths Chapter 8 - Binomial Theorem Formulas  How to Find The Median?  Determinant to Find the Area of a Triangle  Binomial Theorem for Positive Integral Indices  Fourier Series  Taylor Series  The Language of Mathematics  The Converse of Pythagoras Theorem  