JEE Main 2023 — Sequences And Series Question with Solution

From the given series :

$$10=1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\dots$$

We isolate the 1 to get :

$$9=\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\dots$$ .......(1)

Divide each term in the equation by $k$ :

$$\frac{9}{k}=\frac{4}{k^2}+\frac{8}{k^3}+\frac{13}{k^4}+\dots$$ .......(2)

Subtracting the second equation from the first, we obtain :

$$9\left(1-\frac{1}{k}\right)=\frac{4}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\frac{6}{k^4}+\dots$$

This is equivalent to :

$$S=9\left(1-\frac{1}{k}\right)=\frac{4}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\frac{6}{k^4}+\dots$$ .........(3)

Divide both sides by $k$ again, we get : $$\frac{S}{k}=\frac{4}{k^2}+\frac{4}{k^3}+\frac{5}{k^4}+\dots$$ ..........(4)

Subtracting equation (4) from (3), we have :

$$(1-\frac{1}{k})S=\frac{4}{k}+\frac{1}{k^3}+\frac{1}{k^4}+\dots$$

In this equation, you've treated the infinite series on the right-hand side as a geometric series with a ratio of $1/k$, so the sum of this series can be expressed as $\frac{1/k^3}{1-1/k}$.

Therefore, we get :

$$9{\left(1-\frac{1}{k}\right)}^2=\frac{4}{k}+\frac{1/k^3}{1-1/k}.$$

Now, simplifying this equation, we get :

$$9(k-1{)}^2=4k^2-4k+1$$

$$\Rightarrow$$$$9(k^2-2k+1)=4k^2-4k+1$$

$$\Rightarrow$$$$9k^2-18k+9=4k^2-4k+1$$

$$\Rightarrow$$$$9k^2-4k^2-18k+4k+9-1=0$$

$$\Rightarrow$$$$5k^2-14k+8=0$$

This is a standard form quadratic equation, $a{x}^2+bx+c=0$, where $a=5$, $b=-14$, and $c=8$.

The solutions can be found using the quadratic formula, $k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Substituting the values for $a$, $b$, and $c$, we have :

$$\Rightarrow$$$$k=\frac{14\pm \sqrt{(-14{)}^2-4\times 5\times 8}}{2\times 5}$$

$$\Rightarrow$$$$k=\frac{14\pm \sqrt{196-160}}{10}$$

$$\Rightarrow$$$$k=\frac{14\pm \sqrt{36}}{10}$$

$$\Rightarrow$$$$k=\frac{14\pm 6}{10}$$

So, the solutions are $k=2$ or $k=0.8$. However, since the question mentions $k\in \mathbb{N}$, i.e., $k$ is a natural number, the only valid solution is $k=2$.