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1 votes
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1 +2(1/2) +3(1/4)+ 4(1/8)+ .............. = ?
getting 2 please check

3 Answers

Best answer
9 votes
9 votes

$A = 1+2(\frac{1}{2})+3(\frac{1}{4}) + 4(\frac{1}{8})......$

$A(\frac{1}{2}) = 1(\frac{1}{2})+2(\frac{1}{4})+3(\frac{1}{8}) + 4(\frac{1}{16})......$


Substract the equations,

$A(1-\frac{1}{2}) = 1+\frac{1}{2} + \frac{1}{4} + \frac{1}{8}...$

$A(\frac{1}{2}) = 1+1$

$A = 4$

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3 votes
3 votes

$\begin{align*}
&\Rightarrow y = 1+2\cdot\left ( \frac{1}{2} \right ) + 3\cdot\left ( \frac{1}{4} \right ) + 4\cdot\left ( \frac{1}{8} \right ) + \dots \\
&\Rightarrow y = {\color{red}{\sum_{j\geq1}^{\infty} j\cdot x^{j-1}}} \qquad \left [ x = \frac{1}{2} \right ] \\
&\Rightarrow y = \sum_{j\geq0}^{\infty} (j+1)\cdot x^{j}  \qquad \left [ (j-1) \rightarrow j \right ] \\
&\Rightarrow y = \sum_{j\geq0}^{\infty} j\cdot x^{j} + \sum_{j\geq0}^{\infty} x^{j} \\
&\Rightarrow y = x\cdot {\color{red}{\sum_{j\geq1}^{\infty} j\cdot x^{j-1}}} + \sum_{j\geq0}^{\infty} x^{j} \\
&\Rightarrow y = x\cdot y + \sum_{j\geq0}^{\infty} x^{j} \\
&\Rightarrow y = \left ( 1-x \right )^{-1} \cdot \sum_{j\geq0}^{\infty} x^{j}  \\
&\Rightarrow y = \frac{1}{\left ( 1-x \right )^{2}} = 4 \\
\end{align*}$



$\begin{align*} 
&\Rightarrow y = 1 \cdot x^0 + 2 \cdot x^1 + 3 \cdot x^2 + 4 \cdot x^3 + \dots + \infty \\
&\Rightarrow \int y \cdot dx = x + x^2 + x^3 + x^4 + \dots + \infty \\
&\Rightarrow \int y \cdot dx = x \cdot \left ( 1 + x^1 + x^2 + x^3 + \dots + \infty \right ) \\
&\Rightarrow \int y \cdot dx = \frac{x}{1-x} \\
&\Rightarrow y = \frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{x}{1-x} \right ] \\
&\Rightarrow y = \frac{\left ( 1-x \right ) \cdot 1 - x \cdot \left ( -1 \right )}{\left ( 1-x \right )^2} \\
&\Rightarrow y = \left [ \frac{1}{\left ( 1-x \right )^2} \right ]_{\left \{ x = \frac{1}{2} \right \}} = 4 \\
\end{align*}$

edited by
0 votes
0 votes

It is AGP.( AP multiply with GP)

Sum of AGP is limn→∞ Sn = ab/1–r + dbr/(1–r)2.

Hence ans is 4.

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