The value of -r - 2-1 - 2 - - r - 1r-

The correct option is C e2 ∑_r = 2^∞(r 1)r2r! = #160;∑_r = 2^∞12(r 2)! = 12 [10! + 11! + 12! + #160; #160; #160;∞ ] = 12e ...

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Standard XII

Mathematics

Complex Numbers

The value of ...

Question

The value of $\infty \sum r = 2 \frac{1 + 2 + + (r - 1)}{r!}$

e

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2 e

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\frac{e}{2}

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\frac{3 e}{2}

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Solution

The correct option is C $\frac{e}{2}$
$\infty \sum r = 2 \frac{(r - 1) r}{2 r!} = \infty \sum r = 2 \frac{1}{2 (r - 2)!}$

$= \frac{1}{2} [\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \infty] = \frac{1}{2} e$

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The value of ∞∑r=21+2++(r−1)r!

The correct option is C e2∞∑r=2(r−1)r2r!=∞∑r=212(r−2)!=12[10!+11!+12!+∞]=12e

The value of $\infty \sum r = 2 \frac{1 + 2 + + (r - 1)}{r!}$

The correct option is C $\frac{e}{2}$
$\infty \sum r = 2 \frac{(r - 1) r}{2 r!} = \infty \sum r = 2 \frac{1}{2 (r - 2)!}$

$= \frac{1}{2} [\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \infty] = \frac{1}{2} e$