If $| x | < 1$ , then the co-efficient of $x^{n}$ in the expansion of ${(1 + x + x^{2} + x^{3} + . . . . . .)}^{2}$ 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 $n + 1$
Let $S = {(1 + x + x^{2} + x^{3} + . . . .)}^{2}$

$1 + x + x^{2} + . . .$ is in G.P

$∴ S_{n} = \frac{1}{1 - x} = {(1 - x)}^{- 1}$

Now, $S = {[{(1 - x)}^{- 1}]}^{2}$

$= {(1 - x)}^{- 2}$

$= 1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . + (n + 1) x^{n} + . . .$

$∴$ co-efficient of $x^{n}$ is $n + 1$

Suggest Corrections

Similar questions

x^{k} (0 \leq k \leq n)

E = 1 + (1 + x) + (1 + x)^{2} + . . . . + (1 + x)^{n}

| x | < 1

x^{n}

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

| x | < 1

x^{n}

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

x^{n}

(1 + x)^{2 n}

x^{n}

(1 + x)^{2 n - 1}

x

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

Join BYJU'S Learning Program