# Evaluate the summation of 25 times 0.3 to the n plus 1 power, from n equals 2 to 10.

**Solution:**

\( \\\sum_{2}^{10}25(0.3)^{n+1} \\ \\It\: can\: be\: written\: as \\ \\=25\sum_{2}^{10}(0.3)^{n+1}\)

By further calculation

= 25 [(0.3)^{2} + (0.3)^{3} + (0.3)^{4} + (0.3)^{5} + (0.3)^{6} + (0.3)^{7} + (0.3)^{8} + (0.3)^{9} + (0.3)^{10} + (0.3)^{11}]

So we get

= 25 (0.3)^{3} [1 + 0.3^{1} + 0.3^{2} + 0.3^{3} + 0.3^{4} + 0.3^{5} + 0.3^{6} + 0.3^{7} + 0.3^{8}]

The series inside the parenthesis is a geometric series where the common ratio is 0.3 and the first term is 1.

Sum of geometric series can be written as

\( \\S_{n}=a\times (\frac{1-r^{n}}{1-r}) \\ \\Substituting\: the\: values \\ \\S_{9}=1\times (\frac{1-0.3^{9}}{1-0.3})\)So we get

S_{9} = 1.4285

We get

= 25 × (0.3)^{3} × 1.4285

= 0.9642375

**Therefore, the summation is 0.9642375.**

## Evaluate the summation of 25 times 0.3 to the n plus 1 power, from n equals 2 to 10.

**Summary:**

The summation of 25 times 0.3 to the n plus 1 power, from n equals 2 to 10 is 0.9642375.

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