1-21-1-n-31-1-n2- terms - n-0 equals

The correct option is B n^2 let x=1 1n now let S=1+2x+3x^2+4x^3+......... now multiply S with x now Sx=x+2x^2+3x^3+4x^4......... now ...

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Byju's Answer

Standard XII

Mathematics

Selecting Consecutive Terms in GP

1+21-1/n+31-1...

Question

$1 + 2 (1 - 1 / n) + 3 {(1 - 1 / n)}^{2} + . . . . . . . . . \infty$ terms ; $(n > 0)$ equals

n (1 - 1 / n)

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n^{2}

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n {(1 - 1 / n)}^{2}

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N o n e o f t h e s e

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Solution

The correct option is B $n^{2}$
let $x = 1 - \frac{1}{n}$
now let $S = 1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . . .$
now multiply S with $x$
now $S x = x + 2 x^{2} + 3 x^{3} + 4 x^{4} . . . . . . . . .$
now $S - S x = 1 + x + x^{2} + x^{3} + . . . . . . .$
$S (1 - x) = 1 + x + x^{2} + x^{3} + . . . .$
now this series is in GP with common ratio $r = x$
sum of series in GP is $\frac{a}{1 - r}$
$S (1 - x) = \frac{1}{1 - x}$
$S = \frac{1}{(1 - x)^{2}}$
$S = \frac{1}{(1 - (1 - \frac{1}{n}))^{2}} = n^{2}$

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1+2(1−1/n)+3(1−1/n)2+.........∞ terms ; (n>0) equals

The correct option is B n2let x=1−1nnow let S=1+2x+3x2+4x3+.........now multiply S with xnow Sx=x+2x2+3x3+4x4.........now S−Sx=1+x+x2+x3+.......S(1−x)=1+x+x2+x3+....now this series is in GP with common ratio r=xsum of series in GP is a1−rS(1−x)=11−xS=1(1−x)2S=1(1−(1−1n))2=n2

$1 + 2 (1 - 1 / n) + 3 {(1 - 1 / n)}^{2} + . . . . . . . . . \infty$ terms ; $(n > 0)$ equals