Let "r" and "n" be positive integers such that 1 is greater than equal to "r" and "r" is greater than equal to "n". Then price that
C(n,r) + 2.C(n,r-1) + C(n,r-2) = C(n+2,r)

Question

Let "r" and "n" be positive integers such that 1 is greater than
equal to "r" and "r" is greater than equal to "n" Then price
that
C(n,r) + 2 C(n,r-1) + C(n,r-2) = C(n+2,r)

Ayush Jha · Accepted Answer

Since we know that,

$n_{C_{r}} + n_{C_{r + 1}} = {(n + 1)}_{C_{r + 1}}$

Therefore,

$= n_{C_{r}} + 2 n_{C_{r - 1}} + n_{C_{r - 2}} = n_{C_{r - 1}} + n_{C_{r}} + n_{C_{r - 2}} + n_{C_{r - 1}} = {(n + 1)}_{C_{(r)}} + {(n + 1)}_{C_{(r - 1)}} = {(n + 2)}_{C_{(r)}}$

Hence proved.

Let "r" and "n" be positive integers such that 1 is greater than equal to "r" and "r" is greater than equal to "n". Then price thatC(n,r) + 2.C(n,r-1) + C(n,r-2) = C(n+2,r)

Let "r" and "n" be positive integers such that 1 is greater than equal to "r" and "r" is greater than equal to "n". Then price that
C(n,r) + 2.C(n,r-1) + C(n,r-2) = C(n+2,r)