Let S denotes the sum of series 3-23-4-24- 3-5-26- 3-6-27- 5- Then the of S -1 is

Let S = ∑_r = 1^∞r +22^r +1· r (r+1) ⇒ S= ∑_r = 1^∞2(r +1) r2^r +1· r (r+1) ⇒ S= ∑_r = 1^∞(12^r · r 12^r+1· (r +1)) ⇒ S =lim_n →∞(12^1 × 1 12^2 × 2) ...

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Byju's Answer

Standard XI

Mathematics

Geometric Progression

Let S denotes...

Question

Let $S$ denotes the sum of series $\frac{3}{2^{3}} + \frac{4}{2^{4} \cdot 3} + \frac{5}{2^{6} \cdot 3} + \frac{6}{2^{7} \cdot 5} + \dots + \infty .$ Then the of $S^{- 1}$ is

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Solution

Let
$S = \infty \sum r = 1 \frac{r + 2}{2^{r + 1} \cdot r (r + 1)} \Rightarrow S = \infty \sum r = 1 \frac{2 (r + 1) - r}{2^{r + 1} \cdot r (r + 1)} \Rightarrow S = \infty \sum r = 1 (\frac{1}{2^{r} \cdot r} - \frac{1}{2^{r + 1} \cdot (r + 1)}) \Rightarrow S = lim n \to \infty (\frac{1}{2^{1} \times 1} - \frac{1}{2^{2} \times 2}) + (\frac{1}{2^{2} \times 2} - \frac{1}{2^{3} \times 3}) + (\frac{1}{2^{3} \times 3} - \frac{1}{2^{4} \times 4}) + (\frac{1}{2^{n} \times n} - \frac{1}{2^{n + 1} \times (n + 1)}) \Rightarrow S = lim n \to \infty (\frac{1}{2} - \frac{1}{2^{n + 1} \times (n + 1)}) \Rightarrow S = \frac{1}{2} \Rightarrow S^{- 1} = 2$

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Similar questions

Q. Let

S

denotes the sum of series

\frac{3}{2^{3}} + \frac{4}{2^{4} \cdot 3} + \frac{5}{2^{6} \cdot 3} + \frac{6}{2^{7} \cdot 5} + \dots + \infty .

Then the of

S^{- 1}

Q. The value of

\frac{3}{2^{3}} + \frac{4}{2^{4} \cdot 3} + \frac{5}{2^{6} \cdot 3} + \frac{6}{2^{7} \cdot 5} + \dots \infty

is equal to

Q. Let

S

denote the sum of the infinite series

1 + \frac{8}{2!} + \frac{21}{3!} + \frac{40}{4!} + \frac{65}{5!} + . . . . . . .

. Then

Q. Let

S

denote the sum of the infinite series

1 + \frac{8}{2!} + \frac{21}{3!} + \frac{40}{4!} + \frac{65}{5!} + \dots

. Then,

Q. The sum of the series

1 + 2 \cdot 2 + 3 \cdot 2^{2} + 4 \cdot 2^{3} + 5 \cdot 2^{4} + \dots + 100 \cdot 2^{99}

, is

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Let S denotes the sum of series 323+424⋅3+526⋅3+627⋅5+⋯+∞. Then the of S−1 is

Let S=∞∑r=1r+22r+1⋅r(r+1)⇒S=∞∑r=12(r+1)−r2r+1⋅r(r+1)⇒S=∞∑r=1(12r⋅r−12r+1⋅(r+1))⇒S=limn→∞(121×1−122×2) +(122×2−123×3) +(123×3−124×4) +(12n×n−12n+1×(n+1))⇒S=limn→∞(12−12n+1×(n+1))⇒S=12⇒S−1=2

Let $S$ denotes the sum of series $\frac{3}{2^{3}} + \frac{4}{2^{4} \cdot 3} + \frac{5}{2^{6} \cdot 3} + \frac{6}{2^{7} \cdot 5} + \dots + \infty .$ Then the of $S^{- 1}$ is