Question

For Sigma-summation Underscript n = 1 Overscript infinity EndScripts StartFraction 0.2 n Over 0.8 EndFraction, find S3=

Answer 1

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Answer 2

The required summation $\sum5/4(1/5)^n$  for n =3 is 0.31. And the Truncation error is  0.0025.

From image  $\sum5/4(1/5)^n$ to be determine for n = 3 where n = (1,∞).

What is a truncation error?

Truncation error is defined as when the difference in approx value to actual value is a truncated error.

Here,
= $\sum5/4(1/5)^n$

For n = 3
$=5/4((1/5)^1+(1/5)^2+(1/5)^3)\\=5/4*0.248$

=0.3125 ≈ 0.31

Now, truncated error
S =0.3125
Truncated error = 0.3125-0.3100
                           =   0.0025

Thus, The required summation $\sum5/4(1/5)^n$  for n =3 is 0.31. And the Truncation error is 0.0025.

Learn more about truncated error here:
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