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plz can u give me solution of q48 and q49

plz can u give me solution of q48 and q49

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4 Answers

Akshay
185 Points
8 years ago
q49. Sn will be equal to coff. of x in : (1+1/x)n.(1+x)n = ∑nCi*nCi+1 (that is power xi+1 from second bracket multiplied by power xi of first, and i varying from 0 to n-1. (1+1/x)n.(1+x)n = (1+x)2n/xn. Hence Sn(coff. of x = 2nCn+1.Put Sn+1 and Sn and you will get a quardatic equation and n=2 or 4.
Akshay
185 Points
8 years ago
q48. Assume ((x+1)2-1)n = ∑ nCi (1+x)2n-2i. cofficient of xn will be the series in question.
Expand (x+1)2 in LHS and solve.
ATB.
Akshay
185 Points
8 years ago
 will be the series in question.n. cofficient of x2n-2i (-1)j(1+x)iCn = ∑ n-1)2q48. Assume ((x+1)
Expand (x+1)2 in LHS and solve.
ATB.
Akshay
185 Points
8 years ago
Add (-1)^j is summation term.

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