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1) Evaluate the summation of 25 times 0.3 to the n plus 1 power, from n equals 2 to 10.. 0.005 0.321 0.964 10.714
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\[\large 25\sum_{n=2}^{n=10}0.3^{(n+1)}\] The 25 can be taken outside the sigma sign, the reason being it is a common factor of the 9 terms in the summation. We have a geometric sequence, with r = 0.3, n = 9 and the first term being 0.3^3 = 0.027. The sum of the 9 terms of the sequence is given by: \[\large S _{9}=\frac{0.027(1-0.3^{9})}{1-0.3}\] When you have calculated the sum of the geometric sequence, multiply by 25 to find the required summation.
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