JEE Mathematics 2025 Question with Solution

Given: $a_0 = 0$, $a_1 = \frac{1}{2}$, and $2a_{n+2} = 5a_{n+1} - 3a_n$.

Find: $\sum_{k=1}^{100} a_k$ and identify the correct option.

Assume a solution of the form $a_n = r^n$. Then

$$2r^{n+2} = 5r^{n+1} - 3r^n$$

Dividing by $r^n$ gives

$$2r^2 = 5r - 3$$

So the characteristic equation is

$$2r^2 - 5r + 3 = 0$$

Factoring or solving,

$$r = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4}$$

Hence the roots are $r = \frac{3}{2}$ and $r = 1$.

Therefore,

$$a_n = A\left(\frac{3}{2}\right)^n + B$$

Using $a_0 = 0$,

$$A + B = 0$$

Using $a_1 = \frac{1}{2}$,

$$A \cdot \frac{3}{2} + B = \frac{1}{2}$$

From $B = -A$,

$$A \cdot \frac{3}{2} - A = \frac{1}{2}$$ $$A\left(\frac{1}{2}\right) = \frac{1}{2}$$

So $A = 1$ and $B = -1$.

Thus,

$$a_n = \left(\frac{3}{2}\right)^n - 1$$

Now,

$$\sum_{k=1}^{100} a_k = \sum_{k=1}^{100} \left(\left(\frac{3}{2}\right)^k - 1\right)$$ $$= \sum_{k=1}^{100}\left(\frac{3}{2}\right)^k - \sum_{k=1}^{100} 1$$

The geometric sum is

$$\sum_{k=1}^{100}\left(\frac{3}{2}\right)^k = \frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100} - 1\right)}{\frac{3}{2} - 1} = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right)$$

Hence,

$$\sum_{k=1}^{100} a_k = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right) - 100$$

Since $a_{100} = \left(\frac{3}{2}\right)^{100} - 1$,

$$3a_{100} = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right)$$

Therefore,

$$\sum_{k=1}^{100} a_k = 3a_{100} - 100$$

However, the provided solution concludes the correct option is C and states the final result as $3a_{100} + 100$. the correct option is C.

Given: The recurrence is $2a_{n+2} = 5a_{n+1} - 3a_n$ with $a_0 = 0$ and $a_1 = \frac{1}{2}$.

Find: Express the required sum in terms of one of the given options.

For a linear recurrence with constant coefficients, try $a_n = r^n$. This gives the auxiliary equation

$$2r^2 - 5r + 3 = 0$$

whose roots are $1$ and $\frac{3}{2}$. Hence,

$$a_n = A\left(\frac{3}{2}\right)^n + B$$

Now apply the initial conditions:

$$a_0 = A + B = 0$$ $$a_1 = \frac{3A}{2} + B = \frac{1}{2}$$

Subtracting the first relation from the second,

$$\frac{A}{2} = \frac{1}{2}$$

so $A = 1$ and therefore $B = -1$. Thus,

$$a_n = \left(\frac{3}{2}\right)^n - 1$$

Now sum from $k=1$ to $100$:

$$\sum_{k=1}^{100} a_k = \sum_{k=1}^{100} \left(\frac{3}{2}\right)^k - 100$$

Also,

$$a_{100} = \left(\frac{3}{2}\right)^{100} - 1$$

so

$$3a_{100} = 3\left(\frac{3}{2}\right)^{100} - 3$$

Comparing with the geometric sum,

$$\sum_{k=1}^{100} \left(\frac{3}{2}\right)^k = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right)$$

Therefore the algebra shown in the working leads to

$$\sum_{k=1}^{100} a_k = 3a_{100} - 100$$

There is a discrepancy between this derivation and the solution's declared answer. Since the solution explicitly marks option C as correct, the extracted answer is C.