N-lim - 1- 1-3 - 1-32-1-33- - 1-3n - -

The correct option is C 3/2 lim _ n→∞ #160; [ 1+ 1 3 + 1 3 ^ 2 +...+ 1 3 ^ n #160; ] #160; lim _ n→∞ #160; [ 1( 1 ( 1 3 #160 ...

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Byju's Answer

Standard XII

Mathematics

Global Maxima

n→∞lim [ 1+ 1...

Question

$lim n \to \infty [1 + \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . . . . . . . \frac{1}{3^{n}}] =$

\frac{2}{3}

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\frac{3}{2}

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\frac{1}{3}

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\frac{1}{2}

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Solution

The correct option is C $\frac{3}{2}$
$lim n \to \infty [1 + \frac{1}{3} + \frac{1}{3^{2}} + . . . + \frac{1}{3^{n}}]$

$lim n \to \infty ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1 (1 - {(\frac{1}{3})}^{n + 1})}{1 - \frac{1}{3}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = \frac{3}{2} (1 - \frac{1}{3^{\infty}})$

$= \frac{3}{2} (1 - 0)$
$= \frac{3}{2}$

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limn→∞[1+13+132+133+........13n]=

The correct option is C 32limn→∞[1+13+132+...+13n]limn→∞⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1(1−(13)n+1)1−13⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=32(1−13∞)=32(1−0)=32

$lim n \to \infty [1 + \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . . . . . . . \frac{1}{3^{n}}] =$