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Question

# If S=4−9x+16x2−25x3+36x4−49x5+......∞. Find the value of S for x=−12.

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Solution

## This series is not AP,GP,HP and AGP But we see 4,9,16,25.......∞ are square numbers and 1,(−x),(−x)2,(−x)3,(−x)4,.......∞ is a geometric progression. if we subtract square numbers. We can form an AP. So, multiply given series by common ratio and then subtract it with original series. Let s=4−9x+16x2−25x3+36x4−49x5+......∞ -----(1) Multiply by common difference in equation 1 −xs=−4x+9x2−16x3+25x4−36x5+......∞ -----(2) ) On subtracting equation 2 from equation 1 We get, s(1+x)=4−5x+7x2−9x3+11x4−13x5+......∞ -----(3) Multiply by common difference in equation 3 −s(1+x)x=−4x+5x2−7x3+9x4−11x5+......∞ -----(4) On subtracting equation 4 from equation 3 s(1+x)2=4−x+2x2−2x3+2x4−2x5+......∞ s(1+x)2=4−x+2(x2−x3+x4−x5+......∞) s(1+x)2=4−x+2x2(1−x+x2−......∞) s(1+x)2=4−x+2x2.11+x s(1+x)2=4−x+2x21+x=4+3x+x2x+1 s=x2+3x+4(x+1)3 When x=-12 s=(−12)2+3(−12)+4(−12+1)3 14−32+413=22

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