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Binomial Theorem question

2025 · 22 Jan · Shift 1 · Q47

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Binomial Theorem
2025 · 22 Jan · Shift 1 · Q47

Binomial Theorem question

2025 · 22 Jan · Shift 1 · Q47

JEE MainMathematicsBinomial TheoremNumerical+4 / −1

If

\sum_{r=0}^5 \frac{{ }^{11} C_{2 r+1}}{2 r+2}=\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1

, then

\mathrm{m}-\mathrm{n}

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 2035

We need to evaluate

S=\sum_{r=0}^5 \frac{\binom{11}{2r+1}}{2r+2}.

Use the identity

\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}.

Here, take $n=11$ and $k=2r+1$ . Then

\frac{\binom{11}{2r+1}}{2r+2}=\frac{1}{12}\binom{12}{2r+2}.

So

S=\frac{1}{12}\sum_{r=0}^5 \binom{12}{2r+2}.

As $r$ goes from $0$ to $5$ , the index $2r+2$ takes values

2,4,6,8,10,12.

Hence

S=\frac{1}{12}\left(\binom{12}{2}+\binom{12}{4}+\binom{12}{6}+\binom{12}{8}+\binom{12}{10}+\binom{12}{12}\right).

Now use the binomial identity: sum of all even-indexed binomial coefficients is

\sum_{k\text{ even}} \binom{12}{k}=2^{11}=2048.

Thus,

\binom{12}{0}+\binom{12}{2}+\binom{12}{4}+\binom{12}{6}+\binom{12}{8}+\binom{12}{10}+\binom{12}{12}=2048.

Subtracting $\binom{12}{0}=1$ ,

\binom{12}{2}+\binom{12}{4}+\binom{12}{6}+\binom{12}{8}+\binom{12}{10}+\binom{12}{12}=2047.

Therefore,

S=\frac{2047}{12}.

So in lowest terms,

m=2047,\qquad n=12.

Since $2047$ is not divisible by $2$ or $3$ , $\gcd(2047,12)=1$ .

Hence,

m-n=2047-12=2035.

Therefore, the required integer is

\boxed{2035}.

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