$\mathbb E[(\frac{X+1}{4}-\theta)^2]=?$

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Let $\mathbb E[X]=\theta$

What will be $\mathbb E[(\frac{X+1}{4}-\theta)^2]=?$

I found it $$\mathbb E[(\frac{X+1}{4}-\theta)^2]=(5/8)\theta^2-(1/4)\theta+(1/16).$$

But the result is $$\mathbb E[(\frac{X+1}{4}-\theta)^2]=(1/8)\theta^2-(1/8)\theta+(1/16).$$

EDIT:

$X\sim \text{binomial}(2,\theta)$. So

$\mathbb E[X^2]=\mathbb Var[X]+(E[X])^2=2\theta(1-\theta)+4\theta^2$.

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I have done the following $$\mathbb{E}\bigg[\Big(\frac{X+1}{4}-\theta\Big)^2\bigg] = \mathbb{E}\bigg[\Big(\frac{X+1-4\theta}{4}\Big)^2\bigg] = \mathbb{E}\bigg[\frac{16\theta^2-8X\theta-8\theta+X^2+2X+1}{16}\bigg] =$$ $$ = \frac{1}{16}\bigg(\mathbb{E}[16\theta^2] + \mathbb{E}[-8X\theta] + \mathbb{E}[-8\theta] + \mathbb{E}[X^2] + \mathbb{E}[2X]+\mathbb{E}[1]\bigg) =$$ $$ = \frac{1}{16}\bigg(16\theta^2 -8\theta\mathbb{E}[X] -8\theta + \mathbb{E}[X^2] + 2\mathbb{E}[X]+1\bigg) =$$ $$ = \theta^2 -\frac{\theta\mathbb{E}[X]}{2} - \frac{\theta}{2} +\frac{\mathbb{E}[X^2]}{16} + \frac{\mathbb{E}[X]}{8} + \frac{1}{16} = $$ $$ = \theta^2 -\frac{\theta^2}{2} - \frac{\theta}{2} +\frac{\mathbb{E}[X^2]}{16} + \frac{\theta}{8} + \frac{1}{16} = $$ $$ = \frac{\theta^2}{2} - \frac{3\theta}{8} + \frac{\mathbb{E}[X^2]}{16} + \frac{1}{16}$$

In order to get your result, it's necessary to have $\mathbb{E}[X^2] = 4\theta - 6\theta^2$.