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JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Binomial Theorem

If

k = 1 \sum 31 (^{31} C_{k}) (^{31} C_{k - 1}) - k = 1 \sum 30 (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)},

Where

α \in R

, then the value of

16 α

is equal to

(A)1411✓ Correct
(B)1320
(C)1615
(D)1855

Answer: (A)

Solution

Using the identity

\sum_{r = 0}^{n}^{n} C_{r} \cdot^{n} C_{n - r} =^{2 n} C_{n}

, we can rewrite the first sum:

k = 1 \sum 31^{31} C_{k} \cdot^{31} C_{k - 1} = k = 1 \sum 31^{31} C_{k} \cdot^{31} C_{31 - (k - 1)} =^{62} C_{30}

Similarly,

\sum_{k = 1}^{30}^{30} C_{k} \cdot^{30} C_{k - 1} =^{60} C_{29}

.

Now compute:

^{62} C_{30} -^{60} C_{29} = \frac{62 !}{30 ! 32 !} - \frac{60 !}{29 ! 31 !} = \frac{60 !}{29 ! 31 !} (\frac{62 \cdot 61}{30 \cdot 32} - 1) = \frac{60 !}{30 ! 31 !} \cdot \frac{2822}{32}

Therefore

α = \frac{2822}{32}

, and

16 α = 16 \times \frac{2822}{32} = 1411

.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option

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