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Grade: 12th pass
```
1^2 C1+ 2^2 C2+ 3^2 C3+ .........n^2 Cn= ? Answer is n(n+1)2^n-2

```
one year ago

## Answers : (2)

Arun
25333 Points
```							Dear Nishi it is∑ r^2 Cr = ∑ r * n (n-1Cr-1) now n∑ (r-1 +1) (n-1Cr-1) Now it is easy task Hope it helps
```
one year ago
Aditya Gupta
2069 Points
```							hello nishi, we know that(1+x)^n= nC0x^0+nC1x^1+nC2x^2+.........+nCnx^ndifferentiate wrt xn(1+x)^(n-1)= nC1 + 2*nC2*x + 3*nC3*x^2 + ….....+ n*nCn*x^(n-1)multiply both sides by xnx(1+x)^(n-1)= nC1*x + 2*nC2*x^2 + 3*nC3*x^3 + ….....+ n*nCn*x^nnow differentiate again wrt xn(x+1)^(n-2)*(nx+1)= nc1+2^2*nC2*x + 3^2*nC3*x^2 + …............ + n^2*nCn*x^(n-1)now put x= 1 on both sides,1^2 C1+ 2^2 C2+ 3^2 C3+ .........n^2 Cn= n*2^(n-2)*(n+1)or n(n+1)*2^(n-2)kindly approve :)
```
one year ago
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