For $n > 2$ , let $C_{r} = {^{n} C}_{r}$ , and $f (n) = 3 C_{2} + 5 C_{4} + 7 C_{6} + . . . . .$ , find $f (6)$ equals

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Solution

$C_{0} + C_{2} x^{2} + C_{4} x^{4} + . . . . . = \frac{1}{2} [{(1 + x)}^{n} + {(1 - x)}^{n}] M u l t i p l y B y x o n b o t h s i d e s w e g e t \Rightarrow C_{0} x + C_{2} x^{3} + . . . . . = \frac{1}{2} [{(1 + x)}^{n} x + {(1 - x)}^{n} x] D i f f e r e n t i a t i n g b o t h s i d e s w i t h r e s p e c t t o x \Rightarrow C_{0} + {3 C}_{2} x^{2} + {5 C}_{4} x^{4} . . . . . = \frac{1}{2} [{(1 + x)}^{n} + x n (1 + x)^{n - 1} + {(1 - x)}^{n} + x n (- 1) (1 - x)^{n - 1}] P u t t i n g x = 1 i n t h e a b o v e e q u a t i o n w e g e t C_{0} + 3 C_{2} + 5 C_{4} + 7 C_{6} + . . . . = \frac{1}{2} [{(2)}^{n} + n (2)^{n - 1} + {(0)}^{n} + n (- 1) (0)^{n - 1}] = 2^{n - 1} + n (2^{n - 2}) (1)_f (n) = 3 C_{2} + 5 C_{4} + 7 C_{6} + . . . . . . . . U s i n g (1) S o f (n) = 2^{n - 1} + n (2^{n - 2}) - 1 f (6) = 2^{6 - 1} + 6 (2^{6 - 2}) - 1 = 32 + 96 - 1 = 127$

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Similar questions

n \geq 2

C_{r} = (\begin{matrix} n r \end{matrix})

a_{n} = \sum_{r = 0}^{n} \frac{1}{C_{r}}

n \in N

S (n) = \sum_{r = 0}^{n} {(- 1)}^{r} \frac{1}{({^{n} C}_{r})}

S = \sum_{r = 0}^{n} {(- 1)}^{r} \frac{({^{r + 2} C}_{r})}{({^{n} C}_{r})}

(2 \leq r \leq n)

{^{n} C}_{r} + 2 {^{n} C}_{r + 1} + {^{n} C}_{r + 2}

\frac{1}{2} C_{1} - \frac{2}{3} C_{2} + \frac{3}{4} C_{3} - \frac{4}{5} C_{4} + . . . . . . + (- 1)^{n + 1} \cdot \frac{n}{n + 1} = \frac{1}{n + 1}

\frac{{^{n} C}_{r} + 4 {^{n} C}_{r + 1} + 6 {^{n} C}_{r + 2} + 4 {^{n} C}_{r + 3} + {^{n} C}_{r + 4}}{{^{n} C}_{r} + 3 {^{n} C}_{r + 1} + 3 {^{n} C}_{r + 2} + {^{n} C}_{r + 3}} = \frac{n + k}{r + k}

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