Sequence with plenty of binomial coefficients.

freemind

Sequence with plenty of binomial coefficients.

**Posted:** Sat Mar 01, 2008 4:46 pm

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AYMANE

Using $f(x) = \sum_{0}^{p}\frac {C_{p}^{k} x^{k + 4}}{(k + 2)(k + 4)}$ then $a_p = f( - 1)$
I found $a_p = \frac {p^2 + 7p + 6}{2(p + 1)(p + 2)(p + 3)(p + 4)} = \frac {1}{p + 2} - \frac {3}{2}\frac {1}{p + 3} +\frac{1}{2}\frac...$
$\sum_{n = 0}^{ + \infty}a_n = \frac {1}{3}$
Please check it Razz

**Posted:** Fri Mar 28, 2008 2:22 pm

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