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

Question

Let $$ be a sequence such that $$a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$$. If $$28 \sum_\limits{k=1}^{10} \frac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}$$, where $$\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}$$ are the first $$\mathrm{m}$$ prime numbers, then $$\mathrm{m}$$ is equal to

Accepted Answer

Given the sum of the first n terms, $S_n = \frac{n^2+3n}{(n+1)(n+2)}$, we can find the nth term $a_n$ as the difference between the sum of the first n terms and the sum of the first n-1 terms :

So,
$a_n = S_n - S_{n-1}$

Solving, we get :

$$a_n = \frac{n^2+3n}{(n+1)(n+2)} - \frac{(n-1)^2+3(n-1)}{n(n+1)}$$

Simplifying further, we find :

$$a_n = \frac{4}{n(n+1)(n+2)}$$

Then, we find the reciprocal of $a_n$:

$$\frac{1}{a_n} = \frac{n(n+1)(n+2)}{4}$$

Now, we sum this over the first 10 terms :

$$\sum\limits_{k=1}^{10} \frac{1}{a_k} = \sum\limits_{k=1}^{10} \frac{k(k+1)(k+2)}{4}$$

Evaluating the sum :

$$
\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} \left[\sum\limits_{k=1}^{10} k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)\right]
$$

This can be rewritten as the sum of differences :

$$
\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} [(1 \cdot 2 \cdot 3 \cdot 4 - 0) + (2 \cdot 3 \cdot 4 \cdot 5 - 1 \cdot 2 \cdot 3 \cdot 4) + \cdots + (10 \cdot 11 \cdot 12 \cdot 13 - 9 \cdot 10 \cdot 11 \cdot 12)]
$$

$$
\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} (10 \cdot 11 \cdot 12 \cdot 13 - 0) = \frac{1}{16} \cdot 17160
$$

Now, given the condition that :

$$
28 \sum\limits_{k=1}^{10} \frac{1}{a_k} = p_1p_2p_3...p_m
$$

Substituting the sum we've calculated:

$$
28 \cdot \frac{1}{16} \cdot 17160 = p_1p_2p_3...p_m
$$

This simplifies to :

$30030 = p_1p_2p_3...p_m$

The prime factorization of 30030 is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$, which consists of 6 primes.

Therefore, m is equal to 6.

Solutions · 1