If 1-xn-r-0n ar xr - Then 1-a1-a01-a2-a1 -1-an-an-1 is equal toA- n-1n-1-n -B- nn-1-n-1 -C- n-1n-1-n-1 -D- n-1n-n -

The correct option is D (n+1)^nn !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 ...

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Standard XII

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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Binomial Expression

Standard XII 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