Question

If the largest possible coefficient of $x^{2016}$ in $(1 + x^{3n} + x^{504})^{10}$ (where $n \leq 25, n \in N$) is c, then the value of n + c is

• A. $^{10}C_4 + ^4P_4$
• B. $^{10}C_3 + ^6P_6$
• C. $^{11}C_4 + ^6P_6$
• D. $^{11}C_4 + ^4P_4$

Answer 1

Answer: (D)

$^{11}C_4 + ^4P_4$

Answer 2

The general term in the expansion of $(1 + x^{3n} + x^{504})^{10}$ is given by the multinomial theorem: $\frac{10!}{p!q!r!} (1)^p (x^{3n})^q (x^{504})^r = \frac{10!}{p!q!r!} x^{3nq + 504r}$ where $p, q, r$ are non-negative integers such that $p+q+r = 10$.

We are looking for the coefficient of $x^{2016}$, so we must have: $3nq + 504r = 2016$ Dividing the equation by 3, we get: $nq + 168r = 672$ We are given that $n$ is a natural number and $n \leq 25$, so $1 \leq n \leq 25$. Also, $q$ and $r$ are non-negative integers, and $q+r \leq 10$ (since $p = 10-q-r \geq 0$).

From $nq + 168r = 672$, since $nq \geq 0$, we have $168r \leq 672$, which implies $r \leq \frac{672}{168} = 4$. So, possible values for $r$ are $0, 1, 2, 3, 4$.

Case 1: $r = 4$ $nq + 168(4) = 672 \implies nq = 0$. Since $n \geq 1$, $q$ must be 0. The condition $q+r \leq 10$ becomes $0+4 \leq 10$, which is satisfied. For $q=0, r=4$, we get $p = 10 - 0 - 4 = 6$. The triplet $(p,q,r) = (6,0,4)$ is valid for any $n$ in the range $1 \leq n \leq 25$. The coefficient for this triplet is $\frac{10!}{6!0!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.

Case 2: $r = 3$ $nq + 168(3) = 672 \implies nq = 168$. The condition $q+r \leq 10$ becomes $q+3 \leq 10$, so $q \leq 7$. We need to find $n$ ($1 \leq n \leq 25$) such that $nq=168$ and $q \leq 7$.

• If $q=1, n=168$ (invalid $n$)
• If $q=2, n=84$ (invalid $n$)
• If $q=3, n=56$ (invalid $n$)
• If $q=4, n=42$ (invalid $n$)
• If $q=5, n=168/5$ (not integer)
• If $q=6, n=28$ (invalid $n$)
• If $q=7, n=168/7 = 24$. This is a valid value for $n$ ($1 \leq 24 \leq 25$). So, for $n=24$, we have $q=7, r=3$. This gives $p = 10 - 7 - 3 = 0$. The triplet $(p,q,r) = (0,7,3)$ is valid only for $n=24$. The coefficient for this triplet is $\frac{10!}{0!7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.

Case 3: $r = 2$ $nq + 168(2) = 672 \implies nq = 336$. We need $q \leq 10-2=8$. If $q=8, n=336/8 = 42$ (invalid $n$). No valid $n$ in the range $1 \le n \le 25$.

Case 4: $r = 1$ $nq + 168(1) = 672 \implies nq = 504$. We need $q \leq 10-1=9$. If $q=9, n=504/9 = 56$ (invalid $n$). No valid $n$ in the range $1 \le n \le 25$.

Case 5: $r = 0$ $nq + 168(0) = 672 \implies nq = 672$. We need $q \leq 10-0=10$. If $q=10, n=67.2$ (not integer). No valid $n$ in the range $1 \le n \le 25$.

For any $n$ such that $1 \leq n \leq 25$ and $n \neq 24$, the only way to form $x^{2016}$ is using the triplet $(p,q,r)=(6,0,4)$, which gives a coefficient of 210. For $n=24$, we have two ways to form $x^{2016}$:

1. Triplet $(p,q,r)=(6,0,4)$, coefficient = 210.
2. Triplet $(p,q,r)=(0,7,3)$, coefficient = 120. The total coefficient for $n=24$ is $210 + 120 = 330$.

The largest possible coefficient $c$ is 330, which occurs when $n=24$. The value of $n+c$ is $24 + 330 = 354$. Checking the options: Option (4) is $^{11}C_4 + ^4P_4 = \frac{11!}{4!7!} + \frac{4!}{0!} = 330 + 24 = 354$. Thus, $c = ^{11}C_4$ and $n = ^4P_4$.