# SPM Additional Mathematics 2019, Paper 2 (Question 2)

Question 2:
Express 2n + 2 – 2n + 1 + 2n – 1 in the form p(2n – 1), where p is a constant.
Hence, solve the equation 8(2n + 2 – 2n + 1 + 2n – 1) = $5\left({2}^{{n}^{2}}\right).$
[6 marks]

Solution:

$\begin{array}{l}{2}^{n+2}-{2}^{n+1}+{2}^{n-1}\\ =\left({2}^{n}×{2}^{2}\right)-\left({2}^{n}×{2}^{1}\right)+\left({2}^{n}÷{2}^{1}\right)\\ ={2}^{n}\left(4-2+\frac{1}{2}\right)\\ ={2}^{n}\left(\frac{5}{2}\right)\\ =5\left(\frac{{2}^{n}}{2}\right)\\ =5\left({2}^{n-1}\right)\\ \therefore p=5\\ \end{array}$