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showing 1-1 of 1 (answers : 1, comments : 0),   sorted newest to the oldest
Question
$1+(1+a)r+(1+a+a^2)r^2+\cdots$ to $n$ terms.
Find the sum of the above series?
 
(0) (0) Comment Answer
numbers
2016-10-29 19:01:34 
Rachita Aggarwal
1 Answer
$1+(1+a)r+(1+a+a^2)r^2+\cdots\\
=\dfrac{1}{1-a}\left[(1-a)+(1-a^2)r+(1-a^3)r^2+\cdots\right]\\
=\dfrac{1}{1-a}\left[(1+r+r^2+\cdots)-(a+a^2r+a^3r^2+\cdots)\right]\\
=\dfrac{1}{1-a}\left[\dfrac{1-r^n}{1-r}-\dfrac{a(1-a^nr^n)}{1-ar}\right]\\
(\text{where }a\ne1; r\ne 1; ar\ne 1)$
 
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2016-10-29 19:46:30 
Javed Khan (Senior Math Expert, careerbless.com)
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