plz can u give me solution of q48 and q49

q49. Sn will be equal to coff. of x in : (1+1/x)ⁿ.(1+x)ⁿ = ∑ⁿC_i*ⁿC_i+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 = ²ⁿC_n+1.Put S_n+1 and S_n and you will get a quardatic equation and n=2 or 4.

q48. Assume ((x+1)²-1)ⁿ = ∑ ⁿC_i (1+x)^2n-2i. cofficient of xⁿ will be the series in question.

Expand (x+1)² in LHS and solve.

ATB.

Expand (x+1)2 in LHS and solve.

ATB.

Add (-1)^j is summation term.