If the coefficients of three consecutive terms in the expansion of $(1 + x)^n$ are in the ratio $1 : 5 : 20$ , then the coefficient of the fourth term is:
A
$5481$
B
$3654$
C
$2436$
D
$1817$

Question

If the coefficients of three consecutive terms in the expansion of (1 + x)^n are in the ratio 1: 5: 20, then the coefficient of the fourth term is:

Accepted Answer

Given: The coefficients of three consecutive terms in the expansion of $(1 + x)^n$ are in the ratio $1 : 5 : 20$ .

Find: The coefficient of the fourth term.

Let the three consecutive coefficients be

\binom{n}{r-1}, \binom{n}{r}, \binom{n}{r+1}

Then,

\binom{n}{r-1} : \binom{n}{r} : \binom{n}{r+1} = 1 : 5 : 20

From

\frac{\binom{n}{r}}{\binom{n}{r-1}} = 5

we get

\frac{n-r+1}{r} = 5

So,

n-r+1 = 5r

n = 6r - 1

Also,

\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{20}{5} = 4

Hence,

\frac{n-r}{r+1} = 4

So,

n-r = 4r + 4

n = 5r + 4

Equating the two values of $n$ ,

6r - 1 = 5r + 4

r = 5

Therefore,

n = 29

The fourth term in the expansion of $(1+x)^n$ has coefficient

\binom{n}{3}

So,

\binom{29}{3} = \frac{29 \times 28 \times 27}{3 \times 2 \times 1} = 3654

Therefore, the coefficient of the fourth term is $3654$ . The correct option is B.

Given: Three consecutive binomial coefficients are in the ratio $1:5:20$ .

Find: The coefficient of the fourth term of $(1+x)^n$ .

For consecutive coefficients,

\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r} \quad \text{and} \quad \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1}

Since the ratio is $1:5:20$ , the first adjacent ratio is $5$ and the second adjacent ratio is $4$ .

Thus,

\frac{n-r+1}{r} = 5, \qquad \frac{n-r}{r+1} = 4

From the first equation,

n = 6r - 1

From the second equation,

n = 5r + 4

Comparing,

6r - 1 = 5r + 4 \implies r = 5

Hence,

n = 29

Now the fourth term of $(1+x)^n$ is

T_4 = \binom{n}{3}x^3

Therefore its coefficient is

\binom{29}{3} = 3654

Therefore, the required coefficient is $3654$ .

Answer