JEE Mathematics 2024 Question with Solution

Given: $\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots$ up to $10$ terms and $\beta = \sum_{n=1}^{10} n^4$.

Find: $k$ from $4\alpha - \beta = 55k + 40$.

Identify the sequence inside $\alpha$: $1, 4, 8, 13, 19, 26, \ldots$. Its first differences are $3, 4, 5, 6, 7, \ldots$, so the general term is quadratic. Let

$$T_n = an^2 + bn + c$$

Using

$$T_1 = 1, \quad T_2 = 4, \quad T_3 = 8$$

we get

$$a + b + c = 1$$ $$4a + 2b + c = 4$$ $$9a + 3b + c = 8$$

Solving,

$$a = \frac{1}{2}, \quad b = \frac{3}{2}, \quad c = -1$$

Hence

$$T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1$$

So

$$\alpha = \sum_{n=1}^{10} \left(\frac{1}{2}n^2 + \frac{3}{2}n - 1\right)^2$$

Therefore,

$$4\alpha = \sum_{n=1}^{10} (n^2 + 3n - 2)^2$$

From the extracted working,

$$\alpha = 11197, \qquad \beta = \sum_{n=1}^{10} n^4 = 25333$$

Thus

$$4\alpha = 4 \cdot 11197 = 44788$$ $$4\alpha - \beta = 44788 - 25333 = 19455$$

Now use

$$19455 = 55k + 40$$ $$55k = 19415$$ $$k = \frac{19415}{55} = 353$$

Therefore, the value of $k$ is $353$.

Given: the terms inside $\alpha$ are $1, 4, 8, 13, 19, 26, \ldots$.

Find: evaluate $\alpha$ and then solve for $k$.

Using the pattern of successive increments, the next terms are

$$34, 43, 53, 64$$

So the first $10$ terms contribute

$$\alpha = 1 + 16 + 64 + 169 + 361 + 676 + 1156 + 1849 + 2809 + 4096 = 11197$$

Also,

$$\beta = \sum_{n=1}^{10} n^4 = 1 + 16 + 81 + 256 + 625 + 1296 + 2401 + 4096 + 6561 + 10000 = 25333$$

Then

$$4\alpha - \beta = 4(11197) - 25333 = 44788 - 25333 = 19455$$

Now solve

$$19455 = 55k + 40$$ $$55k = 19415$$ $$k = 353$$

Therefore, the value of $k$ is $353$.