JEE Main 2022 (Online) 29th July Morning ShiftQuestion 1
If $\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\frac{1}{4\times5\times6}+\ldots+\frac{1}{100\times101\times102}=\frac{k}{101}$, then $34k$ is equal to ________.
Sequence and Series · Telescoping ($V_n$ Method)
Answer$286$
Solution
Formulas used. (1) $\frac1{n(n+1)(n+2)}=\frac12\left[\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right]$(2) Telescoping: $\sum_{n=p}^{q}(V_n-V_{n+1})=V_p-V_{q+1}$Step 1 - Telescope ($n=2$ to $100$, $V_n=\dfrac1{n(n+1)}$): $$S=\frac12\left[\frac1{2\cdot3}-\frac1{101\cdot102}\right]=\frac12\left[\frac16-\frac1{10302}\right].$$ Step 2 - Simplify. $10302=6\times1717$: $$S=\frac12\cdot\frac{1717-1}{10302}=\frac{858}{10302}=\frac{143}{1717}.$$ Step 3 - Find $k$. $1717=17\times101$: $$\frac{k}{101}=\frac{143}{17\times101}\ \Longrightarrow\ k=\frac{143}{17},\qquad 34k=286 .$$ Answer: 286