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Find the sum of the series 1+\frac{15}{8}+\frac{15}{8} \cdot \frac{21}{16}+\ldots. after identifying it as a binomial expansion.

Option: 1

28


Option: 2

32


Option: 3

36


Option: 4

48


Answers (1)

best_answer

Let the given series be identical with

1+n x+\frac{n(n-1}{2 !} x^{2}+\ldots

Comparing this with the given series, we get

\mathrm{nx}=\frac{15}{8} \\                               ............(1)

\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \mathrm{x}^{2}=\frac{15}{8} \cdot \frac{21}{16}   ...........(2)

Dividing (2) by the square of (1), we get

\frac{n(n-1) x^{2}}{2 n^{2} x^{2}}=\frac{15}{8} \cdot \frac{21}{16} \times\left(\frac{8}{15}\right)^{2} \\

or,  \frac{n-1}{2 n}=\frac{7}{10}, or, n=-\frac{5}{2}

From (1), -\frac{5}{2} x=\frac{15}{8}, or x=-\frac{3}{4} 

Hence the given series is the expansion of \left(1-\frac{3}{4}\right)^{-5 / 2}=\left(\frac{1}{4}\right)^{-5 / 2}=4^{5 / 2}=32.

Posted by

rishi.raj

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