I'm doing the OCR MEI further maths Pure 1 2017 exam paper, checked the mark scheme but i still dont understand the working out of the simplification of the expression.
There is one question that i can understand how they simplified the question below to get the target expression.
$$\sum_{i=1}^ {k+1} \frac{r}{2^r}= 2-(\frac{1}{2})^k (2+k)+\frac{k+1}{2^{k+1}}$$
The target is:
$$\sum_{i=1}^ {k+1}\frac{r}{2^r}=2-(\frac{1}{2})^{k+1}(3+k) $$
Thanks you very much.
The simplification of the RHS comes from
\begin{align} 2 - \left(\frac{1}{2}\right)^k \left(2+k\right)+\frac{k+1}{2^{k+1}} &= 2-\left(\frac{1}{2}\right)^{k+1} 2\left(2+k\right)-\left(\frac{1}{2}\right)^{k+1}\left(-k-1\right)\\ &= 2-\left(\frac{1}{2}\right)^{k+1}\left(4 + 2k - k - 1\right) \\ &= 2-\left(\frac{1}{2}\right)^{k+1}\left(3 + k\right) \end{align}
In the first line above, I multiplied the numerator & denominator by $2$ in the second terms and took out a factor of $-\left(\frac{1}{2}\right)^{k+1}$ in the third term. In the second line, I combined & simplified the second & third terms from the line above.