Probability: Find a value to make this equation hold true

Question

Question:

Let X1,...,Xn be independent, identically distributed random variables with common mean and variance. Find the values of c and d that will make the following formula true: $$E[(X_1 + . . . + X_n)^2] = cE[X_1^2] + d(E[X_1])^2$$

Solution:

$$E[(X_1 +...+X_n)^2] = var(X_1 +...+X_n)+(E[X_1 +...+X_n])^2\\ = nvar(X_1)+(nE[X_1])^2 \\= nE[X_1^2] − n(E[X_1])^2 + n^2E[X_1])^2 \\= nE[X_1^2] + n(n − 1)(E[X_1])^2$$ Thus, c = n and d = n(n − 1).

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I don't understand the very first line of the solution, but I understand the algebra that follows it. Can someone please explain how they get the first line?

Thank you

Answer 1

Well, there is this equality $$ var(X) = E[X^2] - (E[X])^2 $$ which you can find in this wiki article. So, $E[X^2] = var(X) + (E[X])^2 $.