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Byju's Answer

Standard XII

Mathematics

Middle Terms

If |x|<1, t...

Question

If $| x | < 1$ , then the coefficient of $x^{n}$ in the expansion of $(1 + x + x^{2} + x^{3} + . . . . . . . .)^{2}$ is

n

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n - 1

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n + 2

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n + 1

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Solution

The correct option is D $n + 1$
${(\frac{1}{1 - x})}^{2}$
$= (1 - x)^{- 2}$
coefficient of $x^{2} = n + 2 - 1$
$C_{2 - 1}$

$^{n + 1} C_{1}$

$= (n + 1)$

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Similar questions

Q. If

| x | < 1

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Q. If

| x | < 1

then the coefficient of

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{(1 + x + x^{2} + x^{3} . . . .)}^{2}

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

The correct option is D n+1(11−x)2=(1−x)−2coefficient of x2=n+2−1 C2−1 n+1C1 =(n+1)

If $| x | < 1$ , then the coefficient of $x^{n}$ in the expansion of $(1 + x + x^{2} + x^{3} + . . . . . . . .)^{2}$ is

The correct option is D $n + 1$
${(\frac{1}{1 - x})}^{2}$
$= (1 - x)^{- 2}$
coefficient of $x^{2} = n + 2 - 1$
$C_{2 - 1}$

$^{n + 1} C_{1}$

$= (n + 1)$