nC0 + n+1C1 + n+2C2 + .... + n+rCr is equal to

Question

nC0 + n+1C1 + n+2C2 + .... + n+rCr is equal to (A) n+rCr (B) n+r+1Cr (C) n+r+1Cr+1 (D) none of these

Tardigrade Admin · Accepted Answer

Given expression (n+1C0 + n+1C1)+ n+2C2+ n+3C3 +.....+ n+rCr [∵ nC0=n+1C0] =(n+2C1+ n+2C2)+ n+3C3 + ....+ n+rCr =(n+3C2+ n+3C3)+....+ n+rCr =(n+4C3+ n+4C4) + ....+ n+rCr = n+rCr-1+ n+rCr= n+r+1Cr

Q. $^{n} C_{0} +^{n + 1} C_{1} +^{n + 2} C_{2} + .... +^{n + r} C_{r}$ is equal to

$^{n + r} C_{r}$

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

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

none of these

Q. nC0​+n+1C1​+n+2C2​+....+n+rCr​ is equal to

n+rCr​

n+r+1Cr​

n+r+1Cr+1​

none of these

Given expression (n+1C0​+n+1C1​)+n+2C2​+n+3C3​+.....+n+rCr​ [∵nC0​=n+1C0​] =(n+2C1​+n+2C2​)+n+3C3​+....+n+rCr​ =(n+3C2​+n+3C3​)+....+n+rCr​ =(n+4C3​+n+4C4​)+....+n+rCr​ =n+rCr−1​+n+rCr​=n+r+1Cr​

Q. $^{n} C_{0} +^{n + 1} C_{1} +^{n + 2} C_{2} + .... +^{n + r} C_{r}$ is equal to

$^{n + r} C_{r}$

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

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