Sum to n terms of the series l+(1+x)+(1+x+x2)+(1+x+x2+x3)+…
A
n1−x−x(1−xn)(1−x)2
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B
n1−x+x(1−xn)(1−x)2
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C
n1−x+x(1+xn)(1−x)2
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D
n1−x+x(1−xn)(1+x)2
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Solution
The correct option is An1−x−x(1−xn)(1−x)2 rth term of the series can be written as tr=1+x+x2+x3+...+xr−1=1−xr1−x Sn=n∑r=1tr=n∑r=11−xr1−x=11−x[(1−x)+(1−x2)+(1−x3)+...+(1−xn)]=11−x[n−(x+x2+x3+...+xn)]=11−x[n−x(1−xn)1−x]