If for positive integers r 1, n 2, the coefficients of the (3r)th and (r + 2)th powers of x in the expansion of (1 + x)2n are equal, then n is equal to

Question

If for positive integers r > 1, n > 2, the coefficients of the (3r)th and (r + 2)th powers of x in the expansion of (1 + x)2n are equal, then n is equal to (A) 2r +1  (B) n = 2r (C) 3r (D) r + 1

Tardigrade Admin · Accepted Answer

3r th term in the expansion of (1+x)2 n = 2 n C 3 r -1 x 3 r -1 and (r+2) th term in the expansion of (1+x) =2 n C r +1 x r +1 Given that the binomial coefficients of (3 r -1) and ( r +2) th terms are equal. Thus 2 n C3 r-1= 2 n Cr+1 ⇒ 3 r -1= r +1 or 2 n=(3 r-1)+(r+1) ⇒ 2 r=2 text or 2 n=4 r ⇒ r =1 or n=2 r But r>1 Therefore, n=2r

Q. If for positive integers $r > 1, n > 2$ , the coefficients of the $(3 r)^{t h}$ and $(r + 2)^{t h}$ powers of x in the expansion of $(1 + x)^{2 n}$ are equal, then n is equal to

$2 r + 1$

n = 2r

3r

$r + 1$

Q. If for positive integers r>1,n>2, the coefficients of the (3r)th and (r+2)th powers of x in the expansion of (1+x)2n are equal, then n is equal to

2r+1

n = 2r

3r

r+1

Q. If for positive integers $r > 1, n > 2$ , the coefficients of the $(3 r)^{t h}$ and $(r + 2)^{t h}$ powers of x in the expansion of $(1 + x)^{2 n}$ are equal, then n is equal to

$2 r + 1$

$r + 1$