JEE Main 2023 — Sequences And Series Question with Solution

Given the sum of the first n terms, $S_n=\frac{n^2+3n}{(n+1)(n+2)}$, we can find the n^th 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 :

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

Evaluating the sum :

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

This can be rewritten as the sum of differences :

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

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

Now, given the condition that :

$$28\sum _{k=1}^{10}\frac{1}{a_k}=p_1{p}_2{p}_3...p_m$$

Substituting the sum we've calculated:

$$28\cdot \frac{1}{16}\cdot 17160=p_1{p}_2{p}_3...p_m$$

This simplifies to :

$$30030=p_1{p}_2{p}_3...p_m$$

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.