- m -1 n - k -1 m - p - k m n C m - m C p - C K -A- 4n-3nB- 3n-2nC- 3n-2nD- 4n-1

The correct option is A ∑_m=1^n^nC_m(∑_k=1^m(∑_p=k^m m!/p!(m p)!.p!/k!(p k)! ) ) =∑_m=1^n^nC_m(∑_k=1^m(∑_p=k^m^m kC_p k )m!/k!(m k)! ) =∑_m=1^n^nC_m (∑_k 1^m2 ...

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Byju's Answer

Standard XI

Mathematics

Sum of coefficients of all terms

∑ m =1 n ∑ k ...

Question

$\sum_{m = 1}^{n} (\sum_{k = 1}^{m} (\sum_{p = k}^{m}^{n} C_{m} .^{m} C_{p} .^{p} C_{k})) =$

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Solution

The correct option is B
$\sum_{m = 1}^{n}^{n} C_{m} (\sum_{k = 1}^{m} (\sum_{p = k}^{m} \frac{m!}{p! (m - p)!} . \frac{p!}{k! (p - k)!}))$
$= \sum_{m = 1}^{n}^{n} C_{m} (\sum_{k = 1}^{m} (\sum_{p = k}^{m}^{m - k} C_{p - k}) \frac{m!}{k! (m - k)!})$
$= \sum_{m = 1}^{n}^{n} C_{m} (\sum_{k - 1}^{m} 2^{m - k} .^{m} C_{k})$
$= \sum_{m = 1}^{n}^{n} C_{m} ((1 + 2)^{m} - 2^{m}) = \sum_{m - 1}^{n} (^{n} C_{m} 3^{m} -^{n} C_{m} 2^{m})$
$= (1 + 3)^{n} - 1 - (1 + 2)^{n} + 1 = 4^{n} - 3^{n}$

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∑nm=1(∑mk=1(∑mp=k nCm.mCp.pCk))=

The correct option is B ∑nm=1nCm(∑mk=1(∑mp=km!p!(m−p)!.p!k!(p−k)!)) =∑nm=1nCm(∑mk=1(∑mp=km−kCp−k)m!k!(m−k)!) =∑nm=1nCm(∑mk−12m−k.mCk) =∑nm=1nCm((1+2)m−2m)=∑nm−1(nCm3m−nCm2m) =(1+3)n−1−(1+2)n+1=4n−3n

$\sum_{m = 1}^{n} (\sum_{k = 1}^{m} (\sum_{p = k}^{m}^{n} C_{m} .^{m} C_{p} .^{p} C_{k})) =$