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Question

Find the sum of 11+x+2x1+x2+4x31+x4+....+2nx2n11+x2n.

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Solution

Let f(x)=11+x+2x1+x2+4x31+x4+.....+2nx2n11+x2n

We see that in every term of f(x), the numerator is the derivative of the denominator.

Taking advantage of this, we integrate f(x)

f(x)dx=ln(1+x)+ln(1+x2)+ln(1+x4)+...+ln(1+x2n)

f(x)dx=ln[(1+x)(1+x2)...(1+x2n)]

Now, we multiply and divide by (1-x) inside the logarithm

f(x)dx=ln[(1x)(1+x)(1+x2)...(1+x2n)(1x)]

Using (a+b)(ab)=a2b2 repetitively in the numerator, we get

f(x)dx=ln(1x2n+1)(1x)=ln(1x2n+1)ln(1x)

Now, differentiating on both sides w.r.t. x, we get

f(x)=1(1x2n+1)(x2n+11)+1(1x)

This is the value of the required sum. If limits are asked, remember to put them at the end, after the differentiation.

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