JEE Mathematics 2024 Question with Solution

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

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

The terms inside $\alpha$ are $1, 4, 8, 13, 19, 26,...$. Their first differences are $3, 4, 5, 6, 7,...$, so the sequence is quadratic. Let the general term be

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

Using the first three terms:

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

So,

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

Solving these equations,

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

Hence,

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

Therefore,

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

So,

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

From the extracted working,

$$\alpha = 11197$$

and

$$\beta = \sum_{n=1}^{10} n^4 = 25333$$

Now,

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

Substitute into the given relation:

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

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

Given: The sequence inside $\alpha$ is $1, 4, 8, 13, 19, 26,...$ and $\beta = \sum_{n=1}^{10} n^4$.

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

Observe the differences:

$$4-1=3, \quad 8-4=4, \quad 13-8=5, \quad 19-13=6, \quad 26-19=7$$

So the next terms are

$$34, 43, 53, 64$$

Hence the first $10$ terms are

$$1, 4, 8, 13, 19, 26, 34, 43, 53, 64$$

Now square and add them:

$$1^2 = 1, \quad 4^2 = 16, \quad 8^2 = 64, \quad 13^2 = 169, \quad 19^2 = 361$$ $$26^2 = 676, \quad 34^2 = 1156, \quad 43^2 = 1849, \quad 53^2 = 2809, \quad 64^2 = 4096$$

Therefore,

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

Next,

$$\beta = \sum_{n=1}^{10} n^4 = 1^4 + 2^4 + 3^4 +... + 10^4$$ $$= 1 + 16 + 81 + 256 + 625 + 1296 + 2401 + 4096 + 6561 + 10000 = 25333$$

Now,

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

So,

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

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