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 + 0.3¹ + 0.3² + 0.3³ + 0.3⁴ + 0.3⁵ + 0.3⁶ + 0.3⁷ + 0.3⁸]

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

S₉ = 1.4285

We get

= 25 × (0.3)³ × 1.4285

= 0.9642375

Therefore, the summation is 0.9642375.

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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.