Let m-n - and gcd2- n-1- If 30- 30- 0 -29- 30- 1 -2- 30- 28 -1- 30- 29 -n- 2m- then n-m is equal to Here- n- k -nCk

Let S= ∑^30_r=0(30 r) ^30C_r =30∑^30_r=0^30C_r ∑^30_r=0r·^30C_r =30×2^30 ∑^30_r=1r·30r·^29C_r 1 =30× 2^30 30·2^29 =30·2^29(2 1) =15·2^30 ∴ n=15 and m=30 ⇒ n+m=4 ...

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Byju's Answer

Standard XIII

Mathematics

Sum of Coefficients of All Terms

Let m,n ∈ℕ an...

Question

Let $m, n \in N$ and $gcd (2, n) = 1.$ If $30 (\begin{matrix} 300 \end{matrix}) + 29 (\begin{matrix} 301 \end{matrix}) + \dots + 2 (\begin{matrix} 3028 \end{matrix}) + 1 (\begin{matrix} 3029 \end{matrix}) = n \cdot 2^{m},$ then $n + m$ is equal to
$(Here (\begin{matrix} n k \end{matrix}) =^{n} C_{k})$

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Solution

Let $S = 30 \sum r = 0 (30 - r)^{30} C_{r}$
$= 30 30 \sum r = 0^{30} C_{r} - 30 \sum r = 0 r \cdot^{30} C_{r}$
$= 30 \times 2^{30} - 30 \sum r = 1 r \cdot \frac{30}{r} \cdot^{29} C_{r - 1}$
$= 30 \times 2^{30} - 30 \cdot 2^{29}$
$= 30 \cdot 2^{29} (2 - 1)$
$= 15 \cdot 2^{30}$
$∴ n = 15$ and $m = 30$
$\Rightarrow n + m = 45$

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Similar questions

Q. Let

m, n \in N

and

gcd (2, n) = 1.

30 (\begin{matrix} 300 \end{matrix}) + 29 (\begin{matrix} 301 \end{matrix}) + \dots + 2 (\begin{matrix} 3028 \end{matrix}) + 1 (\begin{matrix} 3029 \end{matrix}) = n \cdot 2^{m},

then

n + m

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(Here (\begin{matrix} n k \end{matrix}) =^{n} C_{k})

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(\begin{matrix} 300 \end{matrix}) (\begin{matrix} 3010 \end{matrix}) - (\begin{matrix} 301 \end{matrix}) (\begin{matrix} 3011 \end{matrix}) + (\begin{matrix} 302 \end{matrix}) (\begin{matrix} 3012 \end{matrix}) - . . . + (\begin{matrix} 3020 \end{matrix}) (\begin{matrix} 3030 \end{matrix})

is where

(\begin{matrix} n r \end{matrix}) = {^{n} C}_{r}

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If $m \neq n$ and $(m + n)^{- 1} \times (m^{- 1} + n^{- 1}) = m^{x} n^{y}$ , then $x + y$ is equal to

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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)=nCk)

Let S=30∑r=0(30−r) 30Cr =3030∑r=030Cr−30∑r=0r⋅30Cr =30×230−30∑r=1r⋅30r⋅29Cr−1 =30×230−30⋅229 =30⋅229(2−1) =15⋅230 ∴n=15 and m=30 ⇒n+m=45