(6.) The sum of the series 1 + (1 +...
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(6.) The sum of the series 1 + (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + ... to n terms is [n(1 - x) - X(1 - x^)] ® © © 1, [ n(1 – x) – «(1 – x”)] [(n-1)x-(1 – x?)] [n(1 - x) - (1 – xņ)]

JEE/Engineering Exams
Maths
Solution
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( S=1+(1+x)+left(1+x+x^{2}right)+-cdots cdotleft(1+x+x^{2}+m-1right) ) ( f=x=x+n+t ) ( Rightarrow S=n+(n-1) x+(n-2) x^{2}+-cdots+2 x^{n-2}+x^{n-1} ) ( S x=n x+(n-1) x^{2}+ldots--+2 x^{n-1} ) ( +x^{n} ) ( S-S x=n-left(x+x^{2}+----+x^{n-1}right)-x^{n} ) ( x=frac{x-xleft(x^{n-1}-1right)}{x-1}-x^{n} ) ( (1-x)=n-frac{xleft(1-x^{n}right)}{1-x}(text { let } x<1) ) ( S=frac{n}{(1-x)}-frac{xleft(1-x^{n}right)}{(1-x)^{2}} ) ( S=frac{1}{(1-x)^{2}}left(n(1-x)-xleft(1-x^{n}right)right) )
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