N C0-n-1 Cn-n C1n Cn-1-n C2-n-1 Cn-2- - -n Cn-

The correct option is B Each term of the sum is of the for ^nC_r ×^(n+1 r)C_n r=^nC_r ×^n+1 rC_1 =^nC_r × (n+1 r)=^nC_r(n+1) r^nC_r ⇒ sum =∑_r=0^n(n+1)^nC_r ∑ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Binomial Coefficients

n C0×n+1 Cn+n...

Question

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

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

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

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

Open in App

Solution

The correct option is B

Each term of the sum is of the for $^{n} C_{r} \times^{(n + 1 - r)} C_{n - r} =^{n} C_{r} \times^{n + 1 - r} C_{1}$
$=^{n} C_{r} \times (n + 1 - r) =^{n} C_{r} (n + 1) - r^{n} C_{r}$
$\Rightarrow$ sum = $\sum_{r = 0}^{n} (n + 1)^{n} C_{r} - \sum_{r = 0}^{n} r^{n} C_{r}$
$= (n + 1) \sum_{r = 0}^{n}^{n} C_{r} - \sum_{r = 0}^{n} r^{n} C_{r}$
= $(n + 1) 2^{n} - n 2^{n - 1}$
$= 2^{n - 1} (2 n + 2 - n) = 2^{n - 1} (n + 2)$
$[\sum_{r = 0}^{n} r^{n} C_{r} = n \sum_{r = 0}^{n}^{n - 1} C_{r - 1} = n \times 2^{n - 1}]$

Suggest Corrections

Similar questions

Q. If

^{2 n + 1} C_{0} +^{2 n + 1} C_{1} +^{2 n + 1} C_{2} + \dots +^{2 n + 1} C_{n} = 4^{11}

then which of the following is/are correct ?

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

$\frac{^{n} c_{0}}{n}$ + $\frac{^{n} c_{1}}{n + 1}$ + $\frac{^{n} c_{2}}{n + 2}$ +_ _ _ _ _ _ + $\frac{^{n} c_{n}}{2 n}$ =

Q. Let

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

and

\frac{A_{n + 1}}{A_{n}} = \frac{15}{4}

,then

n

equals

{2.}^{n} C_{0} + 2^{2} . \frac{^{n} C_{1}}{2} + 2^{3} . \frac{^{n} C_{2}}{3} + \dots + 2^{n + 1} . \frac{^{n} C_{n}}{n + 1} =

Join BYJU'S Learning Program

nC0×n+1Cn+nC1nCn−1+nC2×n−1Cn−2+.....nCn =

The correct option is B Each term of the sum is of the for nCr×(n+1−r)Cn−r=nCr×n+1−rC1 =nCr×(n+1−r)=nCr(n+1)−rnCr ⇒ sum =∑nr=0(n+1)nCr−∑nr=0rnCr =(n+1)∑nr=0nCr−∑nr=0rnCr = (n+1)2n−n2n−1 =2n−1(2n+2−n)=2n−1(n+2) [∑nr=0rnCr=n∑nr=0n−1Cr−1=n×2n−1]

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