To evaluate the summation of 25 times (0.3)^(n + 1) from n = 2 to n = 10, we can start by rewriting the summation:
S = Σ (from n=2 to n=10) 25 * (0.3)^(n + 1)
We can factor out the constant 25 from the summation:
S = 25 * Σ (from n=2 to n=10) (0.3)^(n + 1)
Next, we need to evaluate the summation Σ (from n=2 to n=10) (0.3)^(n + 1). To do this, we first expand the summation:
This means we will calculate:
- (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}
Calculating each of these:
- (0.3)^3 = 0.027
- (0.3)^4 = 0.0081
- (0.3)^5 = 0.00243
- (0.3)^6 = 0.000729
- (0.3)^7 = 0.0002187
- (0.3)^8 = 0.00006561
- (0.3)^9 = 0.000019683
- (0.3)^{10} = 0.0000059049
- (0.3)^{11} = 0.000001771561
Now, let’s sum these values:
S = 0.027 + 0.0081 + 0.00243 + 0.000729 + 0.0002187 + 0.00006561 + 0.000019683 + 0.0000059049 + 0.000001771561 = 0.03786079
Finally, we multiply this result by 25:
S = 25 * 0.03786079 = 0.94651975
Therefore, the summation of 25 times (0.3)^(n + 1) from n = 2 to n = 10 is approximately 0.9465.