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Mathematics Question on Sum of First n Terms of an AP

The sum of the series 11×225C0+12×325C1+13×425C2+.........+126×2725C25\frac{1}{1\times2} ^{^{25}}C_{0} + \frac{1}{2\times3}^{^{25}}C_{1} + \frac{1}{3\times4} ^{^{25}}C_{2} + ......... + \frac{1}{26\times27} ^{^{25}}C_{25}

A

227−126×27\frac{2^{27}-1}{26\times27}

B

227−2826×27\frac{2^{27}-28}{26\times27}

C

12(226+126  27)\frac{1}{2}\left(\frac{2^{26}+1}{26\,\,27}\right)

D

226−152\frac{2^{26}-1}{52}

Answer

227−2826×27\frac{2^{27}-28}{26\times27}

Explanation

Solution

Given series is, 11×225C0+12×325C1+13×425C2\frac{1}{1 \times 2}{ }^{25} C_{0}+\frac{1}{2 \times 3}{ }^{25} C_{1}+\frac{1}{3 \times 4}{ }^{25} C_{2} +…+126×2725C25+\ldots+\frac{1}{26 \times 27}{ }^{25} C_{25} \because \int_\limits{0}^{x}(1+x)^{25} d x=\int_\limits{0}^{x}\left[{ }^{25} C_{0}+{ }^{25} C_{1} x\right. +25C2x2+…+25C25x25]dx \left.+{ }^{25} C_{2} x^{2}+\ldots+{ }^{25} C_{25} x^{25}\right] d x On integrating w.r.t. xx, taking limits 0 to xx, we get [(1+x)2626]0x \left[\frac{(1+x)^{26}}{26}\right]_{0}^{x} =[25C0x+25C1⋅x22+25C2x33+…+25C25⋅x2626]0x=\left[{ }^{25} C_{0} x+{ }^{25} C_{1} \cdot \frac{x^{2}}{2}+{ }^{25} C_{2} \frac{x^{3}}{3}+\ldots+{ }^{25} C_{25} \cdot \frac{x^{26}}{26}\right]_{0}^{x} \Rightarrow \left\\{\frac{1}{26}(1+x)^{26}-\frac{1}{26}\right\\} =25C0x+25C1⋅x22+…+25C25⋅x2626={ }^{25} C_{0} x+{ }^{25} C_{1} \cdot \frac{x^{2}}{2}+\ldots+{ }^{25} C_{25} \cdot \frac{x^{26}}{26} Again, integrating w.r.t. xx, taking limits 0 to 1 , we get 126∫01[1+x)26−1]dx\left.\frac{1}{26} \int_{0}^{1}[1+x)^{26}-1\right] d x = \int_ \limits{0}^{1}\left[{ }^{25} C_{0} x+{ }^{25} C_{1} \cdot \frac{x^{2}}{2}+\ldots+{ }^{25} C_{25} \frac{x^{26}}{26}\right] d x ⇒126[(1+x)2727−x]01\Rightarrow \frac{1}{26}\left[\frac{(1+x)^{27}}{27}-x\right]_{0}^{1} =[25C0x22+25C1⋅x32×3+…+25C25x2726×27]01=\left[{ }^{25} C_{0} \frac{x^{2}}{2}+{ }^{25} C_{1} \cdot \frac{x^{3}}{2 \times 3}+\ldots+{ }^{25} C_{25} \frac{x^{27}}{26 \times 27}\right]_{0}^{1} \Rightarrow \frac{1}{26}\left\\{\frac{2^{27}}{27}-1-\frac{1}{27}\right\\}=\frac{1}{2}{ }^{25} C_{0}+\frac{1}{2 \times 3} \cdot{ }^{25} C_{1} +…+126×27⋅25C25+\ldots+\frac{1}{26 \times 27} \cdot{ }^{25} C_{25} ∴11×2⋅25C0+12×325C1+13×425C2\therefore \frac{1}{1 \times 2} \cdot{ }^{25} C_{0}+\frac{1}{2 \times 3}{ }^{25} C_{1}+\frac{1}{3 \times 4}{ }^{25} C_{2} +…+126×27⋅25C25=227−2826×27+\ldots+\frac{1}{26 \times 27} \cdot{ }^{25} C_{25}=\frac{2^{27}-28}{26 \times 27}