Let a_n be a sequence such that a_1+a_2++a_n = n^2+3n(n+1)(n+2).

Question

Let $\{a_n\}$ be a sequence such that $a_1+a_2+\cdots+a_n = \dfrac{n^2+3n}{(n+1)(n+2)}$. If $28\displaystyle\sum_{k=1}^{10}\frac{1}{a_k} = p_1p_2p_3\cdots p_m$, where $p_1,p_2,\ldots,p_m$ are the first $m$ prime numbers, then $m$ is equal to:

Accepted Answer

$$a_n = S_n-S_{n-1} = \frac{4}{n(n+1)(n+2)} \implies \frac{1}{a_n} = \frac{n(n+1)(n+2)}{4}.$$

$$28\sum_{k=1}^{10}\frac{1}{a_k} = 7\sum_{k=1}^{10}k(k+1)(k+2).$$

Using the telescoping identity $k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2) = 4k(k+1)(k+2)$:

$$\sum_{k=1}^{10}k(k+1)(k+2) = \frac{10\times11\times12\times13}{4}.$$

$$28\sum\frac{1}{a_k} = 7\times\frac{10\times11\times12\times13}{4} = 10\times11\times3\times13\times7 = 30030 = 2\times3\times5\times7\times11\times13.$$

This is the product of the first $6$ primes, so $m=6$.

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