Question

If ${(1+x+x^2)}^{25}=a_0+a_{1}x+a_2{x}^2+\cdot \cdot \cdot +a_{50}\cdot x^{50}$ then $a_0+a_2+a_4+\cdot \cdot \cdot +a_{50}$ is

• A. even
• B. odd & of the form $3n$.
• C. odd & of the form $(3n-1)$
• D. Odd & of the form $(3n+1)$

Answer 1

Answer: (A)

even

$(1+x+x^2{)}^{25}=a_0+a_{1}x+a_2{x}^2+...a_{50}{x}^{50}$

Substituting $x=1$, we get

$3^{25}=a_0+a_1+a_2+...a_{50}$ ...(i)

Substituting $x=-1$ we get

$1=a_0-a_1+a_2-a_3...+a_{50}$ ...(ii)

Adding $i$ and $ii$, we get

$3^{25}+1=2[a_0+a_2+a_4+...a_{50}]$

Hence

$a_0+a_2+a_4+...a_{50}=\frac{3^{25}+1}{2}$

Since the LHS is a positive integer hence the RHS has to be a positive integer.
Therefore

$3^{25}+1$ will be divisible by $2$.

Hence

$3^{25}+1$ will be even.

And

$3^{2n+1}+1$ is always even and is divisible by $4$.

Hence

$\frac{3^{25}+1}{2}$ will be even.