Question

Let $(1+2 x)^{20}=a_{0}+a_{1} x +a_{2} x^{2}+\ldots +a_{20} x^{20}$. Then, $3 a_{0}+2 a_{1}+3 a_{2}+2 a_{3}+3 a_{4}+2 a_{5}+\ldots+2 a_{19}+3 a_{20}$ equals

• A. $\frac{5 \cdot 3^{20}-3}{2}$
• B. $\frac{5 \cdot 3^{20}+3}{2}$
• C. $\frac{5 \cdot 3^{20}+1}{2}$
• D. $\frac{5 \cdot 3^{20}-1}{2}$

Answer 1

Answer: (C)

$(1+2 x)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots +a_{20} x^{20}$
Put $x=1$
$3^{20}=a_{0}+a_{1}+a_{2}+\ldots+a_{20}$ ...(i)
Put $x=-1$,
$1=a_{0}-a_{1}+a_{2}-a_{3}+\ldots+a_{20}$ ...(ii)
On adding Eqs. (i) and (ii), we get
$\frac{3^{20}+1}{2}=a_{0}+a_{2}+a_{4}+\ldots+a_{20}$
On subtracting Eq. (ii) from Eq. (i), we get
$\frac{3^{20}-1}{2}=a_{1}+a_{3}+a_{5}+\ldots+a_{19}$
Now, we have
$3 a_{0}+2 a_{1}+3 a_{2}+2 a_{3}+\ldots+2 a_{19}+3 a_{20}$
$=3\left(a_{0}+a_{2}+a_{4}+\ldots+a_{20}\right)+2\left(a_{1}+a_{3}+\ldots +a_{19}\right)$
$= 3\left(\frac{3^{20}+1}{2}\right)+2\left(\frac{3^{20}-1}{2}\right)$
$= \frac{5 \cdot 3^{20}+1}{2}$