Let m - n - N and gcd2- n -1 - If 30 begin pmatrix 30 0lend pmatrix -29 pmatrix 30 1lendpmatrix - cdots -2 begin pmatrix - 30 - 28lendpmatrix -1 pmatrix 301129end pmatrix - nlcdot -2- m - then n - m is equal to -Here pmatrix n k e n d p m a t r i x -n Ck-

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 XII

Mathematics

Sum of Coefficients of All Terms

Let m , n ∈ N...

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

is equal to

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

Q. For any positive integers m,n

(w i t h n \geq m)

, let

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

. Prove that

(\begin{matrix} n m \end{matrix}) + (\begin{matrix} n - 1 m \end{matrix}) + (\begin{matrix} n - 2 m \end{matrix}) + . . . . . . . . . . + (\begin{matrix} m m \end{matrix}) = (\begin{matrix} n + 1 m + 1 \end{matrix})

Q. Assertion :

\frac{1}{m!} C_{0} + \frac{n}{(m + 1)!} C_{1} + \frac{n (n - 1)}{(m + 2)!} C_{2} + . . . . . + \frac{n (n - 1) . . .2 .1}{(m + n)!} C_{n} = \frac{(m + n + 1) (m + n + 2) . . . (m + 2 n)}{(m + n)!}

Reason: for

r \geq 0

(\begin{matrix} m r \end{matrix}) C_{0} + (\begin{matrix} m r - 1 \end{matrix}) C_{1} + . . . . . . (\begin{matrix} m 0 \end{matrix}) C_{r} = (\begin{matrix} m + n r \end{matrix})

Q. For nonnegative integers

s

and

r,

let

(\begin{matrix} s r \end{matrix}) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{s!}{r! (s - r)!} & if r \leq s, 0 & if r > s . \end{matrix}

For positive integers

m

and

n,

let

g (m, n) = m + n \sum p = 0 \frac{f (m, n, p)}{(\begin{matrix} n + p p \end{matrix})}

where for any nonnegative integer

p,

f (m, n, p) = p \sum i = 0 (\begin{matrix} m i \end{matrix}) (\begin{matrix} n + i p \end{matrix}) (\begin{matrix} p + n p - i \end{matrix}) .

Then which of the following statements is/are TRUE?

If $A = (\begin{matrix} - 6 & - 1 9 & 1 \end{matrix}) B = (\begin{matrix} 2 & 0 - 3 & 5 \end{matrix})$ and $C = (\begin{matrix} - 1 & 8 7 & - 7 \end{matrix})$
Find A-B+C

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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