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

n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n + 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n + 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $n + 1$
We know that $1 + x + x^{2} + x^{3} + . . . = \frac{1}{1 - x} = {(1 - x)}^{- 1}$
$\Rightarrow {(1 + x + x^{2} + x^{3} + . . .)}^{2} = {(\frac{1}{1 - x})}^{2}$
$= {(1 - x)}^{- 2}$
$= 1 + 2 x + 3 x^{2} + 4 x^{3} + . . . + (n + 1) x^{n} + . . .$

$∴$ Coefficient of $x^{n} = n + 1$

Suggest Corrections

Similar questions

| x | < 1

x^{n}

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

x

(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . + \frac{x^{n}}{n!})

If |x| < 1 then the coefficient of $x^{n}$ in the expansion

of $(1 + x + x^{2} + . . . . . . .)^{2}$ will be

| x | < 1

x^{n}

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

\frac{1}{x}

{(1 + x)}^{n} {(1 + \frac{1}{x})}^{n}

\frac{n!}{[(n - 1)! (n + 1)!]}

\frac{(2 n)!}{[(n - 1)! (n + 1)!]}

\frac{(2 n)!}{(2 n - 1)! (2 n + 1)!}

Join BYJU'S Learning Program