Explanation:

We know that the general formula for the binomial expansion is:

$$(a+b)^n=\sum_{r=0}^{n} \binom{n}{r}a^{n-r}b^r$$

where $\binom{n}{r}$ is the binomial coefficient, given by:

$$\binom{n}{r}=\frac{n!}{r!(n-r)!}$$

Finding the expansion:

Let the binomial expansion be of the form $(a+b)^n$. We are given the first three terms of the expansion as 1, 10, and 40. Using the formula for the binomial expansion and substituting the values of $a$, $b$, and the first three terms, we get the following system of equations:

$$(a+b)^n=1 \Rightarrow a^n+b^n=1$$
$$(a+b)^n=10 \Rightarrow \binom{n}{1}a^{n-1}b^1=10 \Rightarrow n a^{n-1}b=10$$
$$(a+b)^n=40 \Rightarrow \binom{n}{2}a^{n-2}b^2=40 \Rightarrow \frac{n(n-1)}{2}a^{n-2}b^2=40$$

Solving this system of equations, we get:

$$a=\frac{1}{2}$$
$$b=\frac{\sqrt{3}}{2}$$
$$n=4$$

The expansion is:

$$(\frac{1}{2}+\frac{\sqrt{3}}{2})^4=\sum_{r=0}^{4} \binom{4}{r}(\frac{1}{2})^{4-r}(\frac{\sqrt{3}}{2})^r$$

Expanding this using the binomial coefficients, we get:

$$(\frac{1}{2}+\frac{\sqrt{3}}{2})^4=\frac{1}{16}(1+4\sqrt{3}+6+4\sqrt{3}+3)=\frac{1}{16}(10+8\sqrt{3})=\frac{5}{8}+\frac{\sqrt{3}}{2}$$

Hence, the expansion of the binomial is $\frac{5}{8}+\frac{\sqrt{3}}{2}$.

Ans. this question is (1+2)power5