If 1-x-n-a0-a1 x-a2 x2- - -ar xr- then a0-a1-a2-ar is equal toA- nn-1n-2 -n-r-1-r -B- None of the aboveC- n n -1 n -2 - n - r D- n-1n-2 -n-r-r -

The correct option is D (n+1)(n+2)....(n+r)r!we have, (1 x)^ n=a_0+a_1x+a_2x^2+........+a_rx^r+..... and nbsp; (1 x)^ 1=1+x+x^2+x^3+....+x^r+..... Hence, a_0 ...

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Byju's Answer

Standard XII

Mathematics

Binomial Theorem for Any Index

If 1-x-n=a0+a...

Question

If $(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{r} x^{r} + . . . . .$ , then $a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}$ is equal to

\frac{n (n + 1) (n + 2) . . . . (n + r)}{r!}

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\frac{(n + 1) (n + 2) . . . . (n + r)}{r!}

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

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None of the above

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Solution

The correct option is B $\frac{(n + 1) (n + 2) . . . . (n + r)}{r!}$
we have,
$(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{r} x^{r} + . . . . .$
and
$(1 - x)^{- 1} = 1 + x + x^{2} + x^{3} + . . . . + x^{r} + . . . . .$
Hence,
$a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}$
$=$ coefficient of $x^{r}$ in the product of two series
$=$ coefficient of $x^{r}$ in $(1 - x)^{- n} (1 - x)^{- 1}$
$=$ coefficient of $x^{r}$ in $(1 - x)^{- (n + 1)}$
$= \frac{(n + 1) (n + 2) . . . . (n + r)}{r!}$

Alternate Solution:
General term in the expansion of $(1 - x)^{- n}$ will be
$T_{r + 1} =^{n + r - 1} C_{r} x^{r}$
hence $a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}$
$=^{n - 1} C_{0} +^{n} C_{1} +^{n + 1} C_{2} + \dots +^{n + r - 1} C_{r}$
Now as we know that
$^{n - 1} C_{0} +^{n} C_{1} =^{n} C_{0} +^{n} C_{1} =^{n + 1} C_{1}$
similarly $^{n + 1} C_{1} +^{n + 1} C_{2} =^{n + 2} C_{2}$
Hence
$^{n - 1} C_{0} +^{n} C_{1} +^{n + 1} C_{2} + \dots +^{n + r - 1} C_{r} =^{n + r} C_{r} = \frac{(n + r)!}{n! \cdot r!} = \frac{(n + r) (n + r - 1) \dots (n + 2) (n + 1)}{r!}$

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Similar questions

Q. If

(1 + x)^{n} = n \sum r = 0 a_{r} x^{r} .

Then

(1 + \frac{a_{1}}{a_{0}}) (1 + \frac{a_{2}}{a_{1}}) \dots (1 + \frac{a_{n}}{a_{n - 1}})

is equal to

The value of $n \sum r = 0 (- 1)^{r} \frac{^{n} C_{r}}{r + 2}$ is equal to .

Q. Prove that:

\frac{n!}{(n - r)!} = n (n - 1) (n - 2) . . . (n - (r - 1))

Q. Prove that if

n

and

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are positive integers

n^{r} - n {(n - 1)}^{r} + \frac{n (n - 1)}{2!} {(n - 2)}^{r} - \frac{n (n - 1) (n - 2)}{3!} {(n - 3)}^{r} + \dots

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r = n

Q. Prove that:
(i)

\frac{n!}{(n - r)!}

= n (n − 1) (n − 2) ... (n − (r − 1))

(ii)

\frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!} = \frac{(n + 1)!}{r! (n - r + 1)!}

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If (1−x)−n=a0+a1x+a2x2+........+arxr+....., then a0+a1+a2+......+ar is equal to

If $(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{r} x^{r} + . . . . .$ , then $a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}$ is equal to