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 (2^{n^{2}}) .

$5 (2^{n^{2}}) .$
[6 marks]

Solution:

\begin{array}{l} 2^{n + 2} - 2^{n + 1} + 2^{n - 1} \\ = (2^{n} \times 2^{2}) - (2^{n} \times 2^{1}) + (2^{n} \div 2^{1}) \\ = 2^{n} (4 - 2 + \frac{1}{2}) \\ = 2^{n} (\frac{5}{2}) \\ = 5 (\frac{2^{n}}{2}) \\ = 5 (2^{n - 1}) \\ ∴ p = 5 \end{array}

$\begin{array}{l} 2^{n + 2} - 2^{n + 1} + 2^{n - 1} \\ = (2^{n} \times 2^{2}) - (2^{n} \times 2^{1}) + (2^{n} \div 2^{1}) \\ = 2^{n} (4 - 2 + \frac{1}{2}) \\ = 2^{n} (\frac{5}{2}) \\ = 5 (\frac{2^{n}}{2}) \\ = 5 (2^{n - 1}) \\ ∴ p = 5 \end{array}$

\begin{array}{l} 8 (2^{n + 2} - 2^{n + 1} + 2^{n - 1}) = 5 (2^{n^{2}}) \\ 8 (5) (2^{n - 1}) = 5 {(2)}^{n^{2}} \\ 2^{3} \times 2^{n - 1} = 2^{n^{2}} \\ 3 + n - 1 = n^{2} \\ n^{2} - n - 2 = 0 \\ (n + 1) (n - 2) = 0 \\ n = - 1 o r n = 2 \end{array}

$\begin{array}{l} 8 (2^{n + 2} - 2^{n + 1} + 2^{n - 1}) = 5 (2^{n^{2}}) \\ 8 (5) (2^{n - 1}) = 5 {(2)}^{n^{2}} \\ 2^{3} \times 2^{n - 1} = 2^{n^{2}} \\ 3 + n - 1 = n^{2} \\ n^{2} - n - 2 = 0 \\ (n + 1) (n - 2) = 0 \\ n = - 1 o r n = 2 \end{array}$

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