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

If

\frac{^{11} C _{1}}{2} + \frac{^{11} C _{2}}{3} + \dots + \frac{^{11} C _{9}}{10} = \frac{n}{m}

with

g cd (n, m) = 1

, then

n + m

is equal to

Answer: 2041

Solution

Use

\frac{^{11} C _{r}}{r + 1} = \frac{^{12} C _{r + 1}}{12}

.

r = 1 \sum 9 \frac{^{11} C _{r}}{r + 1} = \frac{1}{12} k = 2 \sum 10^{12} C_{k}

Since

k = 0 \sum 12^{12} C_{k} = 2^{12} = 4096

,

k = 2 \sum 10^{12} C_{k} = 4096 -^{12} C_{0} -^{12} C_{1} -^{12} C_{11} -^{12} C_{12} = 4096 - 1 - 12 - 12 - 1 = 4070

Sum = \frac{4070}{12} = \frac{2035}{6}

Since

2035 = 5 \times 11 \times 37

shares no factor with 6,

n = 2035

,

m = 6

.

n + m = 2041

Key confirmed: 2041.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer

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