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JEE Mathematics 2023 Question with Solution

Suppose the coefficients of three consecutive terms in the binomial expansion of (1+2x)n(1 + 2x)^n are in the ratio 2:5:82 : 5 : 8. Then the coefficient of the term which is in the middle of these three terms is:

  • A

    560560

  • B

    11201120

  • C

    16801680

  • D

    22402240

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The coefficients of three consecutive terms in the expansion of (1+2x)n(1 + 2x)^n are in the ratio 2:5:82:5:8.

Find: The coefficient of the middle term among these three consecutive terms.

The general term in the binomial expansion is

Tk=(nk)(2x)k=(nk)2kxkT_k = \binom{n}{k}(2x)^k = \binom{n}{k}2^k x^k

So, the coefficient of xkx^k is

(nk)2k\binom{n}{k}2^k

Let the three consecutive terms be Tr,Tr+1,Tr+2T_r, T_{r+1}, T_{r+2}. Then their coefficients are

(nr)2r,(nr+1)2r+1,(nr+2)2r+2\binom{n}{r}2^r, \binom{n}{r+1}2^{r+1}, \binom{n}{r+2}2^{r+2}

Using the given ratio,

(nr+1)2r+1(nr)2r=52,(nr+2)2r+2(nr+1)2r+1=85\frac{\binom{n}{r+1}2^{r+1}}{\binom{n}{r}2^r} = \frac{5}{2}, \qquad \frac{\binom{n}{r+2}2^{r+2}}{\binom{n}{r+1}2^{r+1}} = \frac{8}{5}

Detailed Algebra

From the first ratio,

(nr+1)(nr)⋅2=52\frac{\binom{n}{r+1}}{\binom{n}{r}} \cdot 2 = \frac{5}{2}

Hence,

(nr+1)(nr)=54\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{5}{4}

Using

(nr+1)(nr)=n−rr+1\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1}

we get

n−rr+1=54\frac{n-r}{r+1} = \frac{5}{4}

So,

4(n−r)=5(r+1)4(n-r) = 5(r+1) 4n−4r=5r+54n - 4r = 5r + 5 4n=9r+54n = 9r + 5

From the second ratio,

(nr+2)(nr+1)⋅2=85\frac{\binom{n}{r+2}}{\binom{n}{r+1}} \cdot 2 = \frac{8}{5}

Hence,

(nr+2)(nr+1)=45\frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{4}{5}

Using

(nr+2)(nr+1)=n−r−1r+2\frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{n-r-1}{r+2}

we get

n−r−1r+2=45\frac{n-r-1}{r+2} = \frac{4}{5}

So,

5(n−r−1)=4(r+2)5(n-r-1) = 4(r+2) 5n−5r−5=4r+85n - 5r - 5 = 4r + 8 5n=9r+135n = 9r + 13

Now solve the system

4n=9r+54n = 9r + 5

and

5n=9r+135n = 9r + 13

Subtracting the first equation from the second,

n=8n = 8

Substituting into 4n=9r+54n = 9r + 5,

32=9r+532 = 9r + 5 9r=279r = 27 r=3r = 3

The middle term is Tr+1=T4T_{r+1} = T_4, so its coefficient is

(84)24=70⋅16=1120\binom{8}{4}2^4 = 70 \cdot 16 = 1120

Therefore, the coefficient of the middle term is 11201120. Hence, the correct option is B.

Common mistakes

  • Using the ratio of terms instead of the ratio of coefficients. The powers of xx are not part of the coefficient here. Only compare (nk)2k\binom{n}{k}2^k.

  • Forgetting the factor of 22 coming from (2x)k(2x)^k. If that factor is omitted, the equations for nn and rr become incorrect. Always write the coefficient explicitly before forming ratios.

  • Using the wrong identity for consecutive binomial coefficients. The correct relations are (nr+1)(nr)=n−rr+1\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1} and (nr+2)(nr+1)=n−r−1r+2\frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{n-r-1}{r+2}.

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