The sum of the series 1 + 3x + 6x^2 + 10x^3 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $S = 1 + 3x + 6x^2 + 10x^3 + \dots \infty$  Multiply by $x$: $xS = x + 3x^2 + 6x^3 + \dots \infty$  Subtracting the two equations: $S(1 - x) =

Vedclass · Accepted Answer

$\frac{1}{(1 - x)^3}$

Vedclass · Answer

(D) Let $S = 1 + 3x + 6x^2 + 10x^3 + \dots \infty$  Multiply by $x$: $xS = x + 3x^2 + 6x^3 + \dots \infty$  Subtracting the two equations: $S(1 - x) = 1 + 2x + 3x^2 + 4x^3 + \dots \infty$  Multiply by $x$ again: $xS(1 - x) = x + 2x^2 + 3x^3 + \dots \infty$  Subtracting again: $S(1 - x) - xS(1 - x) = 1 + x + x^2 + x^3 + \dots \infty$  $S(1 - x)^2 = \frac{1}{1 - x}$  $S = \frac{1}{(1 - x)^3}$

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

Similar Questions

Find the sum of the following series up to $n$ terms:
$0.6 + 0.66 + 0.666 + \dots$

Sum of the series $\frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{80}{81} + \dots$ to $n$ terms is

The sum of the first $20$ terms of the series $5+11+19+29+41+\ldots$ is $..........$.

If $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:

The value of $5^{1/2} \cdot 5^{1/4} \cdot 5^{1/8} \cdots \infty$ is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module