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Question: Find out the sum of the series \( \dfrac{1}{{1 \times 2}}{}^{25}{C_0} + \dfrac{1}{{2 \times 3}...

Find out the sum of the series
11×225C0+12×325C1+13×425C2+............+126×2725C25 a. 227126×27 b. 2272826×27 c. 12(226126×27) d. 226152  \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}} \\\

Explanation

Solution

Hint: - Use(1+x)25=25C0+25C1x+25C2x2+.................+25C25x25{\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 xx

According to Binomial Theorem the expansion of (1+x)25{\left( {1 + x} \right)^{25}}is
(1+x)25=25C0+25C1x+25C2x2+.................+25C25x25...........(1){\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.xxWith limit 0 to xx
0x(1+x)25dx=0x(25C0+25C1x+25C2x2+.................+25C25x25)dx\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, (1+x)=t...................(2)\left( {1 + x} \right) = t...................\left( 2 \right)
So when
x=0t=1+0=1 x=xt=1+x  x = 0 \Rightarrow t = 1 + 0 = 1 \\\ x = x \Rightarrow t = 1 + x \\\
Now, differentiate equation 2 w.r.t.xx
0+dx=dtdx=dt\Rightarrow 0 + dx = dt \Rightarrow dx = dt
Substitute these value in equation 1
11+xt25dt=0x(25C0+25C1x+25C2x2+.................+25C25x25)dx\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 tndt=[tn+1n+1]\int {{t^n}dt = \left[ {\dfrac{{{t^{n + 1}}}}{{n + 1}}} \right]}
[t2626]11+x=[25C0x+25C1x22+25C2x33+.................+25C25x2626]ox\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

[(1+x)2626126]=[25C0x+25C1x22+25C2x33+.................+25C25x2626000] [(1+x)2626126]=[25C0x+25C1x22+25C2x33+.................+25C25x2626]..............(3)  \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.xxFrom limit 0 to 1.
01[(1+x)2626126]dx=01[25C0x+25C1x22+25C2x33+.................+25C25x2626]dx\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, (1+x)=t...................(4)\left( {1 + x} \right) = t...................\left( 4 \right)
So when
x=0t=1+0=1 x=1t=1+1=2  x = 0 \Rightarrow t = 1 + 0 = 1 \\\ x = 1 \Rightarrow t = 1 + 1 = 2 \\\
Now, differentiate equation 4 w.r.t.xx
0+dx=dtdx=dt\Rightarrow 0 + dx = dt \Rightarrow dx = dt
Substitute these value in equation 3
12[t2626126]dt=01[25C0x+25C1x22+25C2x33+.................+25C25x2626]dx\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 tndt=[tn+1n+1]\int {{t^n}dt = \left[ {\dfrac{{{t^{n + 1}}}}{{n + 1}}} \right]}
[t2727×26126]12=[25C0x22+25C1x32×3+25C2x43×4+.................+25C25x2726×27]01\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
[22727×26126(12727×26)]=[25C012+25C112×3+25C213×4+.................+25C25126×27000] [2272827×26]=11×225C0+12×325C1+13×425C2+............+126×2725C25  \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(1+x)n{\left( {1 + x} \right)^n}, then integrate the expansion w.r.t.xx With limit 0 to x, then again integrate w.r.t.xxwith limit 0 to 1, we will get the required answer.