If 1-xn-r-0n ar xr- then 1-a1a01-a2a1 -1-anan-1 is equal to

A_ra_r 1 = ^nC_r ^nC_r 1 = n r+1r Now, nbsp; 1+a_ra_r 1 =1+ n r+1r ⇒ nbsp;1+a_ra_r 1= n+1r So, nbsp; (1+a_1a_0)(1+a_2a_1) …(1+a_na_n 1) nbsp;= nbsp; ...

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Standard XI

Mathematics

Binomial Expression

If 1+xn=∑r=0n...

Question

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

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

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

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\frac{n^{n - 1}}{(n - 1)!}

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

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Solution

The correct option is B $\frac{(n + 1)^{n}}{n!}$
$\frac{a_{r}}{a_{r - 1}} = \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
Now,
$1 + \frac{a_{r}}{a_{r - 1}} = 1 + \frac{n - r + 1}{r} \Rightarrow 1 + \frac{a_{r}}{a_{r - 1}} = \frac{n + 1}{r}$
So,
$(1 + \frac{a_{1}}{a_{0}}) (1 + \frac{a_{2}}{a_{1}}) \dots (1 + \frac{a_{n}}{a_{n - 1}}) = r = n \prod r = 1 (1 + \frac{a_{r}}{a_{r - 1}}) = r = n \prod r = 1 (\frac{n + 1}{r}) = (n + 1)^{n} r = n \prod r = 1 (\frac{1}{r}) = \frac{(n + 1)^{n}}{n!}$

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Similar questions

Q. If

{(1 + x)}^{n} = \sum_{r = 0}^{n} C_{r} x^{r}

, show that

C_{0} + \frac{C_{1}}{2} + . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

a_{0}, a_{1} {, a}_{2}, . . . a_{n}

is an arithmetic progression of positive integers, then to evaluate

\frac{C_{0}}{a_{0}} \pm \frac{C_{1}}{a_{1}} + \frac{C_{2}}{a_{2}} \pm . . . . u p t o (n + 1) t e r m s

, we integrate both the sides of

\sum_{r = 0}^{n} C_{r} x_{r} = {(1 + x)}^{n}

after multiplying by

x^{r}

for an appropriate value of

r

\frac{1}{3} C_{0} + \frac{1}{4} C_{1} + . . . u p t o (n + 1) t e r m s

equal

Q. If

C_{r}

is coefficient of

x^{r}

in expansion of

{(1 + x)}^{n}, n ϵ N, n \geq 4

then

n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}}

is divisible (remainder 0) by
(1+x)

^{n}

, n

ϵ N

, n

\geq 4

c_{r}, x^{r}

n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}} = 0

Q. If

{(1 - x^{3})}^{n} = \sum_{r = 0}^{n} a_{r} x^{r} {(1 - x)}^{3 n - 2 r}

, then find

a_{r}

, where

n \in N

Q. Assertion :If

f (x) = x^{n}

then

f (1) + \frac{f^{'} (1)}{1} + \frac{f^{''} (1)}{2!} + \frac{f^{'''} (1)}{3!} + . . . + \frac{^{n} f (1)}{n!} = 2^{n}

Reason: If

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

, then

n \sum r = 0^{n} C_{r} = 2^{n}

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Binomial Expression

Standard XI Mathematics

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If (1+x)n=n∑r=0arxr, then (1+a1a0)(1+a2a1)…(1+anan−1) is equal to

The correct option is B (n+1)nn!arar−1= nCr nCr−1=n−r+1r Now, 1+arar−1=1+n−r+1r⇒1+arar−1=n+1r So, (1+a1a0)(1+a2a1)…(1+anan−1)=r=n∏r=1(1+arar−1)=r=n∏r=1(n+1r)=(n+1)nr=n∏r=1(1r)=(n+1)nn!

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