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Binomial Theorem question

2024 · 29 Jan · Shift 1 · Q55

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JEE Main
Mathematics
Binomial Theorem
2024 · 29 Jan · Shift 1 · Q55

Binomial Theorem question

2024 · 29 Jan · Shift 1 · Q55

JEE MainMathematicsBinomial TheoremNumerical+4 / −1

\text { If } \frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots+\frac{{ }^{11} C_9}{10}=\frac{n}{m} \text { with } \operatorname{gcd}(n, m)=1 \text {, then } n+m \text { is equal to }

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Numerical answer

View written solutionFree

Correct answer: 2041

We need to evaluate

\sum_{k=1}^{9}\frac{\binom{11}{k}}{k+1}.$$ 2. Use the standard identity $$\frac{\binom{n}{k}}{k+1}=\frac{1}{n+1}\binom{n+1}{k+1}.$$ For $n=11$, $$\frac{\binom{11}{k}}{k+1}=\frac{1}{12}\binom{12}{k+1}.$$ So $$S=\frac{1}{12}\sum_{k=1}^{9}\binom{12}{k+1}.$$ Let $r=k+1$. Then $r=2$ to $10$, hence $$S=\frac{1}{12}\sum_{r=2}^{10}\binom{12}{r}.$$ 3. Now use $$\sum_{r=0}^{12}\binom{12}{r}=2^{12}=4096.$$ Therefore, $$\sum_{r=2}^{10}\binom{12}{r}=4096-\left[\binom{12}{0}+\binom{12}{1}+\binom{12}{11}+\binom{12}{12}\right].$$ Since $$\binom{12}{0}=1,\quad \binom{12}{1}=12,\quad \binom{12}{11}=12,\quad \binom{12}{12}=1,$$ we get $$\sum_{r=2}^{10}\binom{12}{r}=4096-(1+12+12+1)=4096-26=4070.$$ Thus, $$S=\frac{1}{12}\cdot 4070=\frac{4070}{12}=\frac{2035}{6}.$$ 4. Hence, $$\frac{n}{m}=\frac{2035}{6}$$ with $\gcd(2035,6)=1$. So, $$n=2035,\quad m=6.$$ Therefore, $$n+m=2035+6=2041.$$ 5. Comparison with stored answer: Stored correct answer = $2041$. Our derived answer = $2041$. So they agree.

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