The coefficient of $x^{1012}$ in the expansion of $(1 + x^n + x^{253})^{10}$, (where n $\le$ 22 is any positive integer), is :-

Question

Let $ \alpha, \beta, \gamma $ and $ \delta $ be the coefficients of $ x^7, x^5, x^3, x $ respectively in the expansion of $ (x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5, \, x > 1 $. If $ \alpha u + \beta v = 18 $, $ \gamma u + \delta v = 20 $, then $ u + v $ equals:

Answer 1

To find the coefficient of $x^{1012}$ in the expansion of $(1 + x^n + x^{253})^{10}$, we use the multinomial theorem. This theorem provides a way to expand powers of sums of terms and can be applied here as follows:

The general term in the expansion of $(a + b + c)^m$ is given by:

\[ \frac{m!}{p! \, q! \, r!} a^p b^q c^r \]

such that $p + q + r = m$.

Here, $a = 1$, $b = x^n$, $c = x^{253}$, and $m = 10$. We are interested in the term where the powers of $x$ add up to 1012:

\[ q \cdot n + r \cdot 253 = 1012 \]

Also, we must have $p + q + r = 10$.

Now, solving for $q$ and $r$, we look at:

1. $p + q + r = 10$ 2. $q \cdot n + r \cdot 253 = 1012$

From condition 1, $p = 10 - q - r$.

Now solve for $q$ and $r$:

\[ qn + r \cdot 253 = 1012 \]

We try different values of $q$ and substitute back, bearing in mind that $n \leq 22$.

Consider $r = 4$:

\[ q \cdot n + 4 \cdot 253 = 1012 \] \[ q \cdot n + 1012 = 1012 \]

This gives:

\[ qn = 1012 - 1012 = 0 \Rightarrow q \cdot n = 0 \]

This implies $q = 6$ when $n = 22$, satisfying $p + 6 + 4 = 10$. If we try other configurations, no valid $q, r$ are found.

Thus, the relevant term becomes:

\[ \frac{10!}{6! \, 0! \, 4!} (1)^6 (x^n)^0 (x^{253})^4 = \frac{10!}{6! \cdot 4!} \]

This simplifies to:

\[ {^{10}C_4} \]

Therefore, the coefficient of $x^{1012}$ in the expansion is ${^{10}C_4}$, and the correct answer is:

${^{10}C_4}$

Answer 2