NCERT Solutions
Class 11 Maths
Binomial Theorem
Ex.8.2 Q.11
Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.
It is known that (r + 1) th term, (Tr+1), in the binomial expansion of (a + b) n is given by,
Tr+1 = nCr an-r br
Assuming that xn occurs in the (r + 1) th term of the expansion of (1 + x)2n, we obtain
Tr+1 = 2nCr (1)2n - r xr = 2nCr xr
Comparing the indices of x in xn and in Tr + 1, we obtain r = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n is
2nCn = (2n)! ÷ {n! × (2n - n)!} = (2n)! ÷ {n! × n!} = {(2n)! ÷ (n!)2} ………. (1)
Assuming that xn occurs in the (k +1) th term of the expansion (1 + x)2n – 1, we obtain
Tk+1 = 2n-1Ck (1)2n-1-k xk = 2n-1Ck (x)k
Comparing the indices of x in xn and Tk + 1, we obtain k = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n –1 is
2n-1Cn = (2n - 1)! ÷ {n! × (2n - 1 - n)!}
= (2n - 1)! ÷ {n! × (n – 1)!}
= {2n × (2n - 1)! ÷ {2n × n! × (n – 1)!}
= (2n)! ÷ (2 × n! × n!)
= (2n)! ÷ {2 × (n!)2} ………. (2)
From equation (1) and (2), it is observed that
2nCn ÷ 2 = 2n-1Cn
=> 2nCn = 2(2n-1Cn)
Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.
Hence, proved.