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JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Binomial Theorem

Let

^{n} C_{r}

denote the binomial coefficient of

x^{r}

in the expansion of

(1 + x)^{n}

. If

\sum_{k = 0}^{10} (2^{2} + 3 k)^{n} C_{k} = α \cdot 3^{10} + β \cdot 2^{10}, α, β \in R

, then

α + β

is equal to

Answer: 19

Solution

Writing

2^{2} + 3 k = 4 + 3 k

and splitting the sum,

k = 0 \sum 10 (4 + 3 k)^{10} C_{k} = 4 k = 0 \sum 10^{10} C_{k} + 3 k = 1 \sum 10 k^{10} C_{k}

The first sum is

4 \cdot 2^{10} = 4096

. Using

k^{10} C_{k} = 10^{9} C_{k - 1}

, the second is

3 \cdot 10 \cdot 2^{9} = 15360

. Hence

k = 0 \sum 10 (4 + 3 k)^{10} C_{k} = 4096 + 15360 = 19456 = 19 \cdot 2^{10}

Matching this against

α \cdot 3^{10} + β \cdot 2^{10}

gives

α = 0

and

β = 19

, so

α + β = 19

.

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Numerical answer

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