JEE Main 2023 (Online) 12th April Morning ShiftQuestion 1
Let $\langle a_n\rangle$ be a sequence such that $a_1+a_2+\ldots+a_n=\frac{n^{2}+3n}{(n+1)(n+2)}$. If $28\sum_{k=1}^{10}\frac{1}{a_k}=p_1p_2p_3\ldots p_m$, where $p_1,p_2,\ldots,p_m$ are the first $m$ prime numbers, then $m$ is equal to
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $a_n=S_n-S_{n-1}$ for $n\ge2$, $a_1=S_1$(2) $\sum_{k=1}^{n}k(k+1)(k+2)=\frac{n(n+1)(n+2)(n+3)}{4}$Step 1 - Rewrite $S_n$. $(n+1)(n+2)=n^2+3n+2$, so $$S_n=\frac{(n+1)(n+2)-2}{(n+1)(n+2)}=1-\frac2{(n+1)(n+2)}.$$ Step 2 - General term. $$a_n=\frac2{n(n+1)}-\frac2{(n+1)(n+2)}=\frac{2\left[(n+2)-n\right]}{n(n+1)(n+2)}=\frac4{n(n+1)(n+2)}.$$ For $n=1$: $a_1=S_1=\dfrac46$, which matches. Step 3 - The sum. $$28\sum_{k=1}^{10}\frac1{a_k}=28\cdot\frac14\sum_{k=1}^{10}k(k+1)(k+2)=7\cdot\frac{10\cdot11\cdot12\cdot13}{4}=30030 .$$ Step 4 - Factorise. $$30030=2\cdot3\cdot5\cdot7\cdot11\cdot13,$$ the product of the first $6$ primes. So $m=6$. Answer: (B)