Let $\mathrm{m}, \mathrm{n} \in \mathrm{N}$ and $\operatorname{gcd}(2, \mathrm{n})=1$. If $30\binom{30}{0}+29\binom{30}{1}+\ldots .+2\binom{30}{28}+1\binom{30}{29}=\mathrm{n} .2^{\mathrm{m}}$, then $\mathrm{n}+\mathrm{m}$ is equal to ______ $\left(\text{Here }\binom{n}{k}={ }^{n} C_{k}\right)$

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Binomial Theorem

Let

m, n \in N

and

gcd (2, n) = 1

. If

30 (0 30) + 29 (1 30) + \dots . + 2 (28 30) + 1 (29 30) = n . 2^{m}

, then

n + m

is equal to ______

(Here (k n) =^{n} C_{k})

Answer: 45

Let

S = \sum_{k = 0}^{29} (30 - k) (k 30)

.

We can write:

S = 30 k = 0 \sum 29 (k 30) - k = 0 \sum 29 k (k 30)

Using

k (k n) = n (k - 1 n - 1)

k = 0 \sum 29 k (k 30) = 30 k = 1 \sum 29 (k - 1 29) = 30 (2^{29} - 1)

Also,

\sum_{k = 0}^{29} (k 30) = 2^{30} - 1

(all terms except

(30 30)

).

Therefore:

S = 30 (2^{30} - 1) - 30 (2^{29} - 1) = 30 \cdot 2^{29} - 30 + 30 = 30 \cdot 2^{29} = 15 \cdot 2^{30}

n = 15

(odd, gcd(2,15)=1) and

m = 30

.

Thus

n + m = 15 + 30 = 45

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Integer answer