If r=0^5 ^11C2r+12r+2 = mn, gcd(m, n) = 1, then m - n is equal to… — Solution | JEE Mathematics

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If $\sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{m}{n}$ , gcd( $m, n$ ) = $1$ , then $m - n$ is equal to:

Answer

Correct answer:2035

Standard Method

Given:

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

Find: $m-n$ when $S=\frac{m}{n}$ and gcd( $m,n$ ) = $1$ .

Let $k=2r+1$ . Then the sum runs over odd values of $k$ , so

S = \sum_{\text{odd }k} \frac{{}^{11}C_k}{k+1}

Now use integration:

S = \int_0^1 \sum_{\text{odd }k} {}^{11}C_k x^k \, dx

From the binomial expansion,

(1+x)^{11} - (1-x)^{11} = 2\sum_{\text{odd }k} {}^{11}C_k x^k

Hence,

\sum_{\text{odd }k} {}^{11}C_k x^k = \frac{(1+x)^{11}-(1-x)^{11}}{2}

Therefore,

S = \frac{1}{2}\int_0^1 \left((1+x)^{11}-(1-x)^{11}\right) dx

Integrating,

\int (1+x)^{11} dx = \frac{(1+x)^{12}}{12}, \qquad \int (1-x)^{11} dx = -\frac{(1-x)^{12}}{12}

So,

S = \frac{1}{2}\left[\frac{(1+x)^{12}+(1-x)^{12}}{12}\right]_0^1

Substituting the limits,

S = \frac{1}{24}\left[(2^{12}+0^{12})-(1^{12}+1^{12})\right]

S = \frac{1}{24}(4096-2)=\frac{4094}{24}=\frac{2047}{12}

Thus, $m=2047$ and $n=12$ . Therefore,

m-n=2047-12=2035

So the required answer is $2035$ .

Termwise Verification

Given:

\sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2}

Find: $m-n$ .

Evaluate each term:

\frac{{}^{11}C_1}{2}=\frac{11}{2}, \quad \frac{{}^{11}C_3}{4}=\frac{165}{4}, \quad \frac{{}^{11}C_5}{6}=\frac{462}{6}=77

\frac{{}^{11}C_7}{8}=\frac{330}{8}=\frac{165}{4}, \quad \frac{{}^{11}C_9}{10}=\frac{55}{10}=\frac{11}{2}, \quad \frac{{}^{11}C_{11}}{12}=\frac{1}{12}

Therefore,

S=\frac{11}{2}+\frac{165}{4}+77+\frac{165}{4}+\frac{11}{2}+\frac{1}{12}

Taking denominator $12$ ,

S=\frac{66}{12}+\frac{495}{12}+\frac{924}{12}+\frac{495}{12}+\frac{66}{12}+\frac{1}{12}

S=\frac{2047}{12}

So $m=2047$ and $n=12$ , giving

m-n=2035

Therefore, the required answer is $2035$ .

Writing $\frac{1}{k+1}$ incorrectly when converting the sum into an integral. Since $\int_0^1 x^k \, dx = \frac{1}{k+1}$ , the integrand must be $x^k$ , not $x^{k+1}$ . Use the power of $x$ carefully before integrating.
Forgetting that the summation contains only odd binomial terms. Using the full expansion of $(1+x)^{11}$ directly gives both even and odd powers. Instead, isolate odd terms through $(1+x)^{11}-(1-x)^{11}$ .
Making a sign error while integrating $(1-x)^{11}$ . The antiderivative is $-\frac{(1-x)^{12}}{12}$ , so after subtraction the combined expression becomes a sum inside the bracket. Track the negative sign properly.

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