The sum to first 20 terms of the series 1-3 - 1-12 - 1-48 - - is equal to- 1-3 - 1-12 - 1-48 - - - - 20 - - - - -

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The sum to fi...

Question

The sum to first 20 terms of the series $\frac{1}{3} + \frac{1}{12} + \frac{1}{48} + . . .$ is equal to

श्रेणी $\frac{1}{3} + \frac{1}{12} + \frac{1}{48} + . . .$ के प्रथम 20 पदों का योगफल बराबर है

\frac{4^{20} - 1}{3}

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\frac{2^{40} - 1}{9}

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\frac{1}{9} (4 - 4^{- 19})

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\frac{4}{9}

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Solution

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Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Δ

z_{1}

z_{2}

| \frac{1}{z_{1}} + \frac{1}{z_{2}} | = | \frac{1}{z_{1}} - \frac{1}{z_{2}} |

Δ O P Q

$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

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