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Question

# If |x|<1 and |y|<1, find the sum to infinity of the series (x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+... . Or Find the sum to n terms of the series 11.3+13.5+15.7+... .

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Solution

## Let (x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+... On multiplying and dividing by (x - y), we get E=(x2−y2)x−y+x3−y3x−y+x4−y4x−y+ ... = 1x−y[(x2+x3+x4+... ∞)−(y2+y3+y4+...∞)] = 1x−y(x21−x−y21−y) [∵ s∞=a1−r] = 1x−y[x2(1−y)−y2(1−x)(1−x)(1−y)] = 1x−y[x2−x2y−y2+xy2(1−x)(1−y)] = 1x−y[(x2−y2)−xy(x−y)(1−x)(1−y)] = (x−y)(x+y−xy)(x−y)(1−x)(1−y)=x+y−xy(1−x)(1−y) Or Here, nth term of 1+3+5x ... is an=1+(n−1)2 [∵ an=a+(n−1)d] = 2n−1 Similarly, nth term of 3+5+7+ ... is a′n=2n+1 Now, nth term of the given series is Tn=1(2n−1)(2n+1); n=1,2,3, ...,n = 12[12n−1−12n+1] ∴ Required sum = ∑Tn = ∑12[12n−1−12n+1] = 12[(1−13)+(13−15)+(15−17)+...+(12n−1−12n+1)] = 12(1−12n+1) = 12(2n+1−12n+1)=n2n+1 Hence, sum of the given series is n2n+1

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