- 1-2-2-2-3-2r-2-4-2r-3- to - - is equal to- 1-1-r-3A- 1-r-1-r-2- 1-r-1-r-2A- 1-r-1-r-3A-

The correct option is C 1+r(1 r)^2A 1+4 + 9r^2 + 16r^3 ....∞ A_2(4 + 16r^3) + (1+9r^2) 1 + r + (1 91)^ 2 = A 1+r(1 r)^2A ...

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nCr Definitions and Properties

1^2+2^2+3^2r^...

Question

$1^{2} + 2^{2} + 3^{2} r^{2} + 4^{2} r^{3} + . . . . .$ to $\infty$ is equal to

\frac{1 + r}{(1 - r)^{2}} A

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\frac{1 + r}{(1 - r)^{3}} A

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1 (1 - r)^{3} A

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\frac{1 + r}{(1 - r)^{2}}

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Solution

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Similar questions

A = 1 + r^{a} + r^{2 a} + r^{3 a} . . . . . . \infty

B = 1 + r^{b} + r^{2 b} . . . . . . \infty

\frac{a}{b} =

A = 1 + r^{a} + r^{2 a} + r^{3 a} + \dots \infty, a > 0

B = 1 + r^{2 b} + r^{4 b} + r^{6 b} + \dots \infty, b > 0,

| r | < 1,

\frac{a}{b}

\infty \sum r = 1 \frac{1}{(2 π - 1)^{2}} = \frac{π^{2}}{8}

\infty \sum r = 1 \frac{1}{r^{2}}

a = \sum_{r = 1}^{\infty} \frac{1}{r^{2}}, b = \infty \sum r = 1 \frac{1}{(2 r - 1)^{2}}

\frac{3 a}{b}

\sum_{r = 1}^{\infty} \frac{1}{{(2 r - 1)}^{2}} = \frac{π^{2}}{8}

\sum_{r = 1}^{\infty} \frac{1}{r^{2}}

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