Question

Let K be the number of rational terms in the expansion of $(\sqrt{2}+\sqrt[6]{3})^{6144}$. If the coefficient of $x^P (P \in N)$ in the expansion of $\frac{1}{(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})}$ is $a_P$, then $a_K - a_{K+1} - a_{K-1} =$

• A. 1
• B. 0
• C. -2
• D. 2

Hint:

To simplify products like $(1+x)(1+x^2)\dots$, multiply by $(1-x)$ and use difference of squares repeatedly. This helps find coefficients in generating functions.

Answer 1

The Correct Option is C

Step 1: Rational terms in $(\sqrt{2}+\sqrt[6]{3})^{6144$.}
General term: $T_{r+1} = ^{6144}C_r (2^{1/2})^{6144-r} (3^{1/6})^r$
Exponents must be integers: $r$ multiple of lcm(2,6) = 6.
$K = 0,6,12,\dots,6144 \implies K = 1025$

Step 2: Simplify generating function.
$E(x) = \frac{1}{(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})} = \frac{1-x}{1-x^{32}} = (1-x) \sum_{j=0}^\infty x^{32j}$

Step 3: Coefficients $a_P$.
- $a_P = 1$ if $P$ multiple of 32
- $a_P = -1$ if $P-1$ multiple of 32
- $a_P = 0$ otherwise

Step 4: Compute $a_K - a_{K+1 - a_{K-1}$.}
$K=1025$: $a_{1025} = -1$, $a_{1026}=0$, $a_{1024}=1$
$\therefore a_K - a_{K+1} - a_{K-1} = -1 - 0 - 1 = -2$