JAMB - Mathematics (2006 - No. 45)
How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1 - 2\(^8\)?
12
10
9
8
Explanation
3, - 6, +12, -24
a = 3, r = -2
\(Sn = \frac{a(1-r^n)}{1-r}\\
∴1-2^8 = \frac{3(1-(-2^{n}))}{1-(-2)}\\
1-2^8 = \frac{3(1-(-2^{n}))}{3}\)
1 - 2\(^8\) = 1 - (-2)\(^n\)
Since 1 - 2\(^8\) = 1 - (-2)\(^n\)
a = 3, r = -2
\(Sn = \frac{a(1-r^n)}{1-r}\\
∴1-2^8 = \frac{3(1-(-2^{n}))}{1-(-2)}\\
1-2^8 = \frac{3(1-(-2^{n}))}{3}\)
1 - 2\(^8\) = 1 - (-2)\(^n\)
Since 1 - 2\(^8\) = 1 - (-2)\(^n\)
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