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Binomial Theorem question

2025 · 24 Jan · Shift 1 · Q29
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  5. /2025 · 24 Jan · Shift 1 · Q29

Binomial Theorem question

2025 · 24 Jan · Shift 1 · Q29

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
For some neq10\mathrm{n} eq 10neq10, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of (1+x)n+4(1+\mathrm{x})^{\mathrm{n}+4}(1+x)n+4 be in A.P. Then the largest coefficient in the expansion of (1+x)n+4(1+\mathrm{x})^{\mathrm{n}+4}(1+x)n+4 is:
  1. A
    10
  2. B
    35
  3. C
    70
  4. D
    20
View written solutionFree

Correct answer: B

  1. Write the relevant coefficients

In the expansion of

(1+x)n+4,(1+x)^{n+4},(1+x)n+4,

the general term is

Tr+1=(n+4r)xr.T_{r+1} = \binom{n+4}{r}x^r.Tr+1​=(rn+4​)xr.

So:

  • 5th term coefficient =(n+44)= \binom{n+4}{4}=(4n+4​)
  • 6th term coefficient =(n+45)= \binom{n+4}{5}=(5n+4​)
  • 7th term coefficient =(n+46)= \binom{n+4}{6}=(6n+4​)

These are in A.P., so

2(n+45)=(n+44)+(n+46).2\binom{n+4}{5} = \binom{n+4}{4} + \binom{n+4}{6}.2(5n+4​)=(4n+4​)+(6n+4​).
  1. Let N=n+4N=n+4N=n+4

Then the condition becomes

2(N5)=(N4)+(N6).2\binom{N}{5} = \binom{N}{4} + \binom{N}{6}.2(5N​)=(4N​)+(6N​).

Using binomial coefficient ratios:

(N4)=(N5)⋅5N−4,\binom{N}{4} = \binom{N}{5}\cdot \frac{5}{N-4},(4N​)=(5N​)⋅N−45​,

and

(N6)=(N5)⋅N−56.\binom{N}{6} = \binom{N}{5}\cdot \frac{N-5}{6}.(6N​)=(5N​)⋅6N−5​.

Substitute into the A.P. condition:

2(N5)=(N5)⋅5N−4+(N5)⋅N−56.2\binom{N}{5} = \binom{N}{5}\cdot \frac{5}{N-4} + \binom{N}{5}\cdot \frac{N-5}{6}.2(5N​)=(5N​)⋅N−45​+(5N​)⋅6N−5​.

Since (N5)≠0\binom{N}{5} \neq 0(5N​)=0, divide throughout by it:

2=5N−4+N−56.2 = \frac{5}{N-4} + \frac{N-5}{6}.2=N−45​+6N−5​.
  1. Solve for NNN

Multiply by 6(N−4)6(N-4)6(N−4):

12(N−4)=30+(N−5)(N−4).12(N-4) = 30 + (N-5)(N-4).12(N−4)=30+(N−5)(N−4).

Expand:

12N−48=30+N2−9N+20.12N - 48 = 30 + N^2 - 9N + 20.12N−48=30+N2−9N+20. 12N−48=N2−9N+50.12N - 48 = N^2 - 9N + 50.12N−48=N2−9N+50.

Bring all terms to one side:

N2−21N+98=0.N^2 - 21N + 98 = 0.N2−21N+98=0.

Factorize:

(N−14)(N−7)=0.(N-14)(N-7)=0.(N−14)(N−7)=0.

So,

N=14orN=7.N=14 \quad \text{or} \quad N=7.N=14orN=7.

Since N=n+4N=n+4N=n+4, we get

n=10orn=3.n=10 \quad \text{or} \quad n=3.n=10orn=3.

Given n≠10n \neq 10n=10, therefore

n=3.n=3.n=3.

Hence,

N=n+4=7.N=n+4=7.N=n+4=7.
  1. Find the largest coefficient in (1+x)7(1+x)^7(1+x)7

The binomial coefficients of (1+x)7(1+x)^7(1+x)7 are:

1,  7,  21,  35,  35,  21,  7,  1.1,\;7,\;21,\;35,\;35,\;21,\;7,\;1.1,7,21,35,35,21,7,1.

Thus the largest coefficient is

35.35.35.
  1. Check options
  • A: 101010 ❌
  • B: 353535 ✅
  • C: 707070 ❌
  • D: 202020 ❌

So the correct option is B.

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