Given that x-0- the sum -n-1-x-x-1n-1 equals

The correct option is B x + 1 (b) x + 1 ∑_n=1^∞ ( x/x + 1 )^n 1= 1 + ( x/x+1 ) + ( x/x+1 )^2 + ( x/x+1 )3 + ( x/x+1 )^4 + ...... ∞ =1/1 ( x/x + 1 ) [∴It ...

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Standard XII

Mathematics

Sum of Infinite Terms of a GP

Given that x>...

Question

Given that x > 0, the sum $\sum_{n = 1}^{\infty} {(\frac{x}{x + 1})}^{n - 1}$ equals

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x + 1

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$\frac{x}{2 x + 1}$

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$\frac{x + 1}{2 x + 1}$

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Solution

The correct option is B
x + 1

(b) x + 1

$\sum_{n = 1}^{\infty} {(\frac{x}{x + 1})}^{n - 1} = 1 + (\frac{x}{x + 1}) + {(\frac{x}{x + 1})}^{2} + (\frac{x}{x + 1}) 3 + {(\frac{x}{x + 1})}^{4} + . . . . . . \infty$

$= \frac{1}{1 - (\frac{x}{x + 1})}$

$[∴ It is a G.P. with a = 1 and r = (\frac{x}{x + 1})]$

$= \frac{(x + 1)}{(x + 1 - x)}$

$= \frac{(x + 1)}{1} = (x + 1)$

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Similar questions

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Given that x > 0, the sum ∑∞n=1(xx+1)n−1 equals

The correct option is B x + 1 (b) x + 1 ∑∞n=1(xx+1)n−1=1+(xx+1)+(xx+1)2+(xx+1)3+(xx+1)4+......∞ =11−(xx+1) [∴It is a G.P. with a = 1 andr=(xx+1)] =(x+1)(x+1−x) =(x+1)1=(x+1)

Given that x > 0, the sum $\sum_{n = 1}^{\infty} {(\frac{x}{x + 1})}^{n - 1}$ equals