Suppose A and B are the coefficients of $30^{\text {th }}$ and $12^{\text {th }}$ terms respectively in the binomial expansion of $(1+x)^{2 n-1}$. If $2 A=5 B$, then $n$ is equal to:

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Binomial Theorem

Suppose A and B are the coefficients of

3 0^{th}

and

1 2^{th}

terms respectively in the binomial expansion of

(1 + x)^{2 n - 1}

. If

2 A = 5 B

, then

n

is equal to:

Answer: (B)

Write

N = 2 n - 1

. The coefficient of the

r

-th term of

(1 + x)^{N}

(i.e. of

x^{r - 1}

) is

(r - 1 N)

, so

A = (29 N), B = (11 N) .

The condition

2 A = 5 B

means

\frac{A}{B} = \frac{( 29 N )}{( 11 N )} = \frac{5}{2}

.

Use the consecutive-coefficient ratio

\frac{( r + 1 N )}{( r N )} = \frac{N - r}{r + 1}

, taking

r = 11

\frac{( 12 N )}{( 11 N )} = \frac{N - 11}{12} .

Now use the symmetry

(29 N) = (N - 29 N)

. If

N - 29 = 12

, i.e.

N = 41

, then

(29 N) = (12 N)

and the ratio above directly gives

\frac{A}{B}

\frac{A}{B} = \frac{N - 11}{12} = \frac{41 - 11}{12} = \frac{30}{12} = \frac{5}{2},

which is exactly the required ratio. So

N = 41

works.

Since

N = 2 n - 1 = 41

n = 21.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option