Coefficient of $x^{r}$ in $1 + (1 + x) + (1 + x)^{2} + . . . . . . + (1 + x)^{n}$ is

^{n + 3} C_{r}

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

^{n + 1} C_{r + 1}

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

^{n} C_{r}

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

^{(n + 2)} C_{r}

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

Open in App

Solution

The correct option is A $^{n + 3} C_{r}$
Consider given the series, $1 + (1 + x) + {(1 + x)}^{2} + {(1 + x)}^{3} + . . . . . . . . . . . . . . . {(1 + x)}^{n}$

Given series is G.P. which have first term $a = 1$ ,comman ratio $r = 1 + x$ and number of term $= n$

So sum is,

$s_{n} = \frac{a (r^{n} - 1)}{r - 1} = \frac{[{(1 + x)}^{n} - 1]}{1 + x} = {(1 + x)}^{n - 1} - {(1 + x)}^{- n}$

Now rth term of ${(1 + x)}^{n - 1} - {(1 + x)}^{- n}$ is $=^{n - 1} C_{r} x^{r} -^{- n} C_{r} . x^{r} = (^{n - 1} C_{r} -^{- n} C_{r}) x^{r}$

Hence, coefficient of $x^{r} = (^{n - 1} C_{r} -^{- n} C_{r})$

Hence, this is the answer.

Suggest Corrections

Similar questions

x^{r}

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

^{n + 1} C_{r + 1}

^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . +^{r} C_{r} =^{n + 1} C_{r + 1}

$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C r - 2$ =

T_{r + 1}

{(1 + x)}^{n + 1} =

T_{r}

T_{r + 1}

{(1 + x)}^{n}

^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}

^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}

Join BYJU'S Learning Program