JEE 2023 — Mathematics PYQ

Pehle $S$ ko $\frac{1}{k}$ se multiply karke subtract karte hain:

$S=1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\dots$

$\frac{S}{k}=\frac{1}{k}+\frac{4}{k^2}+\frac{8}{k^3}+\dots$

Subtract karne par:

Step 2: Dobara subtract karna

Maana $S_1=1+\frac{3}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\dots$

Ise phir $\frac{1}{k}$ se multiply karke subtract karein:

$S_1=1+\frac{3}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\dots$

$\frac{S_1}{k}=\frac{1}{k}+\frac{3}{k^2}+\frac{4}{k^3}+\dots$

Subtract karne par:

Brackets waali series ek Infinite GP hai jiska sum $\frac{a}{1-r}=\frac{1/k}{1-1/k}=\frac{1}{k-1}$ hota hai.

Isse humein mila: $S_1=\frac{k^2+k-1}{(k-1{)}^2}$

Step 3: $S=10$ solve karna

Equation (i) se:

Step 4: $k$ ki value nikalna

Humein pata hai $k\in \mathbb{N}$ (Natural Number). Equation mein values rakh kar check karte hain:

• Agar $k=2$:

  $9(8)-31(4)+31(2)-10=72-124+62-10=0$

  Yeh satisfy kar raha hai!

Final Answer:

The value of $k$ is $2$.