The sum of the coefficients of three consecutive terms in the bin… — Solution | JEE Mathematics

MCQMediumJEE 2023Simple Applications

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n+2}$ , which are in the ratio $1 : 3 : 5$ , is equal to:

A
$63$
B
$92$
C
$25$
D
$41$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Three consecutive coefficients in the expansion of $(1+x)^{n+2}$ are in the ratio $1:3:5$ .

Find: The sum of these three coefficients.

Let the three consecutive coefficients be

\binom{n+2}{r-1},\ \binom{n+2}{r},\ \binom{n+2}{r+1}

with ratio $1:3:5$ .

From the first two terms,

\frac{\binom{n+2}{r-1}}{\binom{n+2}{r}}=\frac{1}{3}

Using the binomial coefficient ratio,

\frac{\binom{n+2}{r-1}}{\binom{n+2}{r}}=\frac{r}{n-r+3}

So,

\frac{r}{n-r+3}=\frac{1}{3}

which gives

3r=n-r+3

Therefore,

n=4r-3

Continue solving the ratios

From the second and third terms,

\frac{\binom{n+2}{r}}{\binom{n+2}{r+1}}=\frac{3}{5}

Using the binomial coefficient ratio,

\frac{\binom{n+2}{r}}{\binom{n+2}{r+1}}=\frac{r+1}{n+2-r}

So,

\frac{r+1}{n+2-r}=\frac{3}{5}

which gives

5(r+1)=3(n+2-r)

5r+5=3n+6-3r

8r-1=3n

Solve for the required coefficients

Substitute $n=4r-3$ into $8r-1=3n$ :

8r-1=3(4r-3)

8r-1=12r-9

4r=8

r=2

Then,

n=4(2)-3=5

So the expansion is of $(1+x)^7$ and the required coefficients are

\binom{7}{1},\ \binom{7}{2},\ \binom{7}{3}

Hence,

\binom{7}{1}+\binom{7}{2}+\binom{7}{3}=7+21+35=63

Therefore, the sum of the coefficients is $63$ . The solution states option B, but the working clearly gives $63$ , which matches option A. Hence, the correct option is A.

Common mistakes

Taking the consecutive coefficients as $\binom{n+2}{r}, \binom{n+2}{r+1}, \binom{n+2}{r+2}$ and then mismatching the ratio equations. This shifts the indices incorrectly. Define the middle term first and use $\binom{n+2}{r-1}, \binom{n+2}{r}, \binom{n+2}{r+1}$ .
Using the wrong ratio identity for binomial coefficients. For example, writing $\frac{\binom{n+2}{r-1}}{\binom{n+2}{r}}=\frac{n-r+3}{r}$ instead of its reciprocal. This reverses the equation. Carefully simplify factorials before equating to the given ratio.
Stopping after finding $n=5$ and forgetting that the required quantity is the sum of the three coefficients, not the value of $n$ . After finding $n$ and $r$ , substitute into the actual coefficients and add them.

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Step-by-step solution

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