JAMB - Mathematics (1997 - No. 18)
The nth term of a sequence is given \(3^{1 - n}\), find the sum of the first three terms of the sequence.
\(\frac{13}{9}\)
1
\(\frac{1}{3}\)
\(\frac{1}{9}\)
Explanation
Tn = \(3^{1 - n}\)
where n = 1, 2, 3.
\(S_3 = 3^{1 - 1} + 3^{1 - 2} + 3^{1 - 3}\)
= 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\)
= \(\frac{13}{9}\)
where n = 1, 2, 3.
\(S_3 = 3^{1 - 1} + 3^{1 - 2} + 3^{1 - 3}\)
= 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\)
= \(\frac{13}{9}\)
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