Sum the series- 1 - 1-3 - 1-32 - 1-33 - 1-34-

The correct option is A 3/4 We know, S_∞=a1 r From the given series, a=1 ,r= 13 ∴ S_∞=11 ( 13 ) = 11+13 =143 = 34 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Product of Binomial Coefficients

Sum the serie...

Question

Sum the series: $1 - \frac{1}{3} + \frac{1}{3^{2}} - \frac{1}{3^{3}} + \frac{1}{3^{4}} . . . . . . . \infty$

\frac{3}{4}

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\frac{4}{3}

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\frac{2}{3}

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\frac{1}{3}

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Solution

The correct option is A $\frac{3}{4}$
We know, $S_{\infty} = \frac{a}{1 - r}$
From the given series, $a = 1, r = - \frac{1}{3}$
$∴ S_{\infty} = \frac{1}{1 - (- \frac{1}{3})}$
$= \frac{1}{1 + \frac{1}{3}}$

$= \frac{1}{\frac{4}{3}}$

$= \frac{3}{4}$

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Similar questions

Find the sum of the following series to infinity :

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(ii) $8 + 4 \sqrt{2} + 4 + . . . . \infty$

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(iv) 10 - 9 + 8.1 - 7.29 + ...... $\infty$

(v) $\frac{1}{3} + \frac{1}{5^{2}} + \frac{1}{3^{3}} + \frac{1}{5^{4}} + \frac{1}{3^{5}} + \frac{1}{5^{6}} + . . . . . \infty$

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Q. The value of

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Sum the series: 1−13+132−133+134.......∞

The correct option is A 34We know, S∞=a1−rFrom the given series, a=1,r=−13∴S∞=11−(−13)=11+13=143=34

Sum the series: $1 - \frac{1}{3} + \frac{1}{3^{2}} - \frac{1}{3^{3}} + \frac{1}{3^{4}} . . . . . . . \infty$

The correct option is A $\frac{3}{4}$
We know, $S_{\infty} = \frac{a}{1 - r}$
From the given series, $a = 1, r = - \frac{1}{3}$
$∴ S_{\infty} = \frac{1}{1 - (- \frac{1}{3})}$
$= \frac{1}{1 + \frac{1}{3}}$

$= \frac{1}{\frac{4}{3}}$

$= \frac{3}{4}$