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Question Number 54647 by gunawan last updated on 08/Feb/19

show that

a . \underset{r = 1}{\overset{n}{Σ}} r^{3} ._{n} C_{r} = n^{2} (n + 3) . 2^{n - 3}

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

Commented by maxmathsup by imad last updated on 08/Feb/19

l e t s (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} = {(x + 1)}^{n} \Rightarrow s^{^{'}} (x) = \sum_{k = 1}^{n} k C_{n}^{k} x^{k - 1} = n {(x + 1)}^{n - 1} \Rightarrow

\sum_{k = 1}^{n} k C_{n}^{k} x^{k} = n x {(x + 1)}^{n - 1} \Rightarrow \sum_{k = 1}^{n} k^{2} C_{n}^{k} x^{k - 1} = n {(x + 1)}^{n - 1} + n x (n - 1) {(x + 1)}^{n - 2} \Rightarrow

\sum_{k = 1}^{n} k^{2} C_{n}^{k} x^{k} = n x {(x + 1)}^{n - 1} + n (n - 1) x^{2} {(x + 1)}^{n - 2} \Rightarrow

\sum_{k = 1}^{n} k^{3} C_{n}^{k} x^{k - 1} = n {(x + 1)}^{n - 1} + n (n - 1) x {(x + 1)}^{n - 2} + 2 n (n - 1) x {(x + 1)}^{n - 2}

+ n (n - 1) (n - 2) x^{2} {(x + 1)}^{n - 3} \Rightarrow

\sum_{k = 1}^{n} k^{3} C_{n}^{k} x^{k} = n x {(x + 1)}^{n - 1} + n (n - 1) x^{2} {(x + 1)}^{n - 2} + 2 n (n - 1) x^{2} {(x + 1)}^{n - 2}

+ n (n - 1) (n - 2) x^{3} {(x + 1)}^{n - 3} f o r x = 1 w e g e t

\sum_{k = 1}^{n} k^{3} C_{n}^{k} = n 2^{n - 1} + n (n - 1) 2^{n - 2} + 2 n (n - 1) 2^{n - 2} + n (n - 1) (n - 2) 2^{n - 3}

= n 2^{n - 1} + (n^{2} - n + 2 n^{2} - 2 n) 2^{n - 2} + n (n - 1) (n - 2) 2^{n - 3}

= n 2^{n - 1} + (3 n^{2} - 3 n) 2^{n - 2} + n (n - 1) (n - 2) 2^{n - 3}

= n 2^{n - 1} + (6 n^{2} - 6 n) 2^{n - 3} + (n^{2} - n) (n - 2) 2^{n - 3}

= n 2^{n - 1} + (6 n^{2} - 6 n + n^{3} - 3 n^{2} + 2 n) 2^{n - 3}

= n 2^{n - 1} + (n^{3} + 3 n^{2} - 4 n) 2^{n - 3}

= 4 n 2^{n - 3} + (n^{3} + 3 n^{2} - 4 n) 2^{n - 3} = (n^{3} + 3 n^{2}) 2^{n - 3} = n^{2} (n + 3) 2^{n - 3}

a n d t h e r e s u l t i s p r o v e d .

Commented by gunawan last updated on 09/Feb/19

Wow Thank you very much Sir

Commented by Abdo msup. last updated on 09/Feb/19

y o u a r e w e l c o m e s i r .