Let m , n N and gcd ( 2 , n ) = 1 . If 30 ( 30 0 ) + 29 ( 30 1 ) + ? + 2 ( 30 28 ) + ( 30 29 ) = n · 2 m , then n + m is equal to _________. ( Here, ( n k ) = C k n )

Question

Let $m, n \in N$ and $\gcd (2, n) = 1$ . If $30 (\binom{30}{0}) + 29 (\binom{30}{1}) + \dots + 2 (\binom{30}{28}) + 1 (\binom{30}{29}) = n \cdot 2^{m},$ then $n + m$ is equal to _________. $(Here, (\binom{n}{k}) = {}^{n}C_{k})$

Smart Achievers · Accepted Answer

(45)

$We have,$

$30 (\binom{30}{0}) + 29 (\binom{30}{1}) + \dots + 2 (\binom{30}{28}) + 1 (\binom{30}{29})$

$a = 30 (\binom{30}{30}) + 29 (\binom{30}{29}) + \dots + 2 (\binom{30}{2}) + 1 (\binom{30}{1})$

$= \sum_{r = 1}^{30} r (\binom{30}{r}) = \sum_{r = 1}^{30} r \cdot \frac{30}{r} (\binom{29}{r - 1})$

$= 30 [(\binom{29}{0}) + (\binom{29}{1}) + (\binom{29}{2}) + \dots + (\binom{29}{29})]$

$= 30 \cdot 2^{29} = 15 \times 2^{30} = n \cdot 2^{m}$ $(given)$

$∴$ $n + m = 15 + 30 = 45$

Solution

Q.
Let $m, n \in N$ and $\gcd (2, n) = 1$ . If $30 (\binom{30}{0}) + 29 (\binom{30}{1}) + \dots + 2 (\binom{30}{28}) + 1 (\binom{30}{29}) = n \cdot 2^{m},$ then $n + m$ is equal to _________. $(Here, (\binom{n}{k}) = {}^{n}C_{k})$

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Solution

Q. Let m,n∈N and gcd(2,n)=1. If 30(300)+29(301)+⋯+2(3028)+1(3029)=n·2m, then n+m is equal to _________. (Here, (nk)=Ckn)

Ans. (45) We have, 30(300)+29(301)+⋯+2(3028)+1(3029) a=30(3030)+29(3029)+⋯+2(302)+1(301) =∑r=130r(30r)=∑r=130r·30r(29r-1) =30[(290)+(291)+(292)+⋯+(2929)] =30·229=15×230=n·2m (given) ∴ n+m=15+30=45

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Q.
Let $m, n \in N$ and $\gcd (2, n) = 1$ . If $30 (\binom{30}{0}) + 29 (\binom{30}{1}) + \dots + 2 (\binom{30}{28}) + 1 (\binom{30}{29}) = n \cdot 2^{m},$ then $n + m$ is equal to _________. $(Here, (\binom{n}{k}) = {}^{n}C_{k})$