If - x -1- then in the expansion of 1-2 x-3 x2-4 x3- -1-2- the coefficient of xn isA- nB- 1C- n-1D- -1

The correct option is B 1 Since 1 + 2x + 3x^2 + 4x^3 + ....... nbsp;∞ = (1 x)^ 2 Therefore, we have nbsp; (1 + 2x + 3x^2 + 4x^3 + ....... nbsp;∞)^ 1/2 ...

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Byju's Answer

Standard XI

Mathematics

Binomial Theorem

If | x |<1, t...

Question

If |x| < 1, then in the expansion of

$(1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . .)^{\frac{1}{2}}$ , the coefficient of $x^{n}$ is

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Solution

The correct option is C
1

Since $1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . \infty = (1 - x)^{- 2}$

Therefore, we have

$(1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . \infty)^{\frac{1}{2}}$ = $((1 - x)^{- 2})^{\frac{1}{2}}$

= $(1 - x)^{- 1}$ = $1 + x + x^{2} + . . . . . . + x^{n} + . . . . . . . \infty$

∴ The coefficient of $x^{n}$ = 1.

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If |x| < 1, then in the expansion of (1+2x+3x2+4x3+.......)12, the coefficient of xn is

The correct option is C 1 Since 1+2x+3x2+4x3+.......∞=(1−x)−2 Therefore, we have (1+2x+3x2+4x3+.......∞)12 = ((1−x)−2)12 = (1−x)−1 =1+x+x2+......+xn+.......∞ ∴ The coefficient of xn = 1.

If |x| < 1, then in the expansion of

$(1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . .)^{\frac{1}{2}}$ , the coefficient of $x^{n}$ is