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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
2 months ago

Answers : (2)

Arun
22595 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
2 months ago
Aditya Gupta
1673 Points
							
hello nishi, we know that
(1+x)^n= nC0x^0+nC1x^1+nC2x^2+.........+nCnx^n
differentiate wrt x
n(1+x)^(n-1)= nC1 + 2*nC2*x + 3*nC3*x^2 + ….....+ n*nCn*x^(n-1)
multiply both sides by x
nx(1+x)^(n-1)= nC1*x + 2*nC2*x^2 + 3*nC3*x^3 + ….....+ n*nCn*x^n
now differentiate again wrt x
n(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 :)
 
2 months ago
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