Let nCr denote the binomial coefficient of xr in the expansion of 1-xn- If -k-01022-3knCk-310-210- - - then - is equal to

N must be equal to 10 ∑_k=0^10(2^2+3k)^nC_k =∑_k=0^10(4+3k)^10C_k =4∑_k=0^10^10C_k+3∑_k=0^10k ^10C_k = 4(2)^10+3× 10∑_k=0^9^9C_k =4(2^10)+30×2^9 ⇒ nbsp;α· ...

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Standard XII

Mathematics

Multinomial Expansion

Let nCr denot...

Question

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of ${(1 + x)}^{n} .$ If $10 \sum k = 0 (2^{2} + 3 k)^{n} C_{k} = α \cdot 3^{10} + β \cdot 2^{10},$ $α, β \in R,$ then $α + β$ is equal to

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Solution

$n$ must be equal to $10$
$10 \sum k = 0 (2^{2} + 3 k)^{n} C_{k} = 10 \sum k = 0 (4 + 3 k)^{10} C_{k} = 4 10 \sum k = 0^{10} C_{k} + 3 10 \sum k = 0 k^{10} C_{k} = 4 (2)^{10} + 3 \times 10 9 \sum k = 0^{9} C_{k} = 4 (2^{10}) + 30 \times 2^{9} \Rightarrow α \cdot 3^{10} + β \cdot 2^{10} = 19 \times 2^{10} ∴ α + β = 19$

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Let nCr denote the binomial coefficient of xr in the expansion of (1+x)n. If 10∑k=0(22+3k)nCk=α⋅310+β⋅210, α, β∈R, then α+β is equal to

n must be equal to 10 10∑k=0(22+3k)nCk=10∑k=0(4+3k)10Ck=410∑k=010Ck+310∑k=0k 10Ck=4(2)10+3×109∑k=09Ck=4(210)+30×29⇒α⋅310+β⋅210=19×210∴α+β=19

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of ${(1 + x)}^{n} .$ If $10 \sum k = 0 (2^{2} + 3 k)^{n} C_{k} = α \cdot 3^{10} + β \cdot 2^{10},$ $α, β \in R,$ then $α + β$ is equal to