Question
The sum of the co-efficients of all the integral powers of $$x$$ in the expansion of $${\left( {1 + 2\sqrt x } \right)^{40}}$$ is
A.
$${3^{40}} + 1$$
B.
$${3^{40}} - 1$$
C.
$$\frac{1}{2}\left( {{3^{40}} - 1} \right)$$
D.
$$\frac{1}{2}\left( {{3^{40}} + 1} \right)$$
Answer :
$$\frac{1}{2}\left( {{3^{40}} + 1} \right)$$
Solution :
The co-efficients of the integral powers of $$x$$ are $$^{40}{C_0},{\,^{40}}{C_2} \cdot {2^2},{\,^{40}}{C_4} \cdot {2^4},.....,{\,^{40}}{C_{40}} \cdot {2^{40}}.$$
$$\eqalign{
& {\left( {1 + 2} \right)^{40}} = {\,^{40}}{C_0} + {\,^{40}}{C_1} \cdot 2 + {\,^{40}}{C_2} \cdot {2^2} + ..... + {\,^{40}}{C_{40}} \cdot {2^{40}}. \cr
& {\left( {1 - 2} \right)^{40}} = {\,^{40}}{C_0} - {\,^{40}}{C_1} \cdot 2 - {\,^{40}}{C_2} \cdot {2^2} - ..... + {\,^{40}}{C_{40}} \cdot {2^{40}}. \cr} $$
Adding, $${3^{40}} + 1 = 2 \times \left( {{\text{required sum}}} \right).$$