Could anyone please tell me if there's any formula or something similar to calculate the following series easily without using a calculator.
$$1.12+(1.12)^2+(1.12)^3+(1.12)^4+(1.12)^5+(1.12)^6+(1.12)^7+(1.12)^8+(1.12)^9$$
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6This is a Geometric Series – lulu Aug 13 '21 at 17:28
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2If you add dollar signs on each side of the expression, it becomes a mathematical formula – LurchiDerLurch Aug 13 '21 at 17:29
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2The formula you're looking for is the geometric series partial sum formula. It's still not easy to calculate without a calculator but it is easier. – Daniel Aug 13 '21 at 17:30
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@lulu Thank you so much! All the exponents made me confused and I missed it. – Russell Aug 13 '21 at 17:42
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1Related post on MSE here – Arctic Char Aug 13 '21 at 17:43
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If you want to work out the actual value of the result, you’ll need to work out $(1.12)^{10}$. I’d do this with exponentiation by squaring, or do a binomial approximation to a couple of terms if that’s acceptable. – Milten Aug 13 '21 at 17:45
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Let $S_{n}$ is a Geometrics series and $x= \frac{s_{i+1}}{s_{i}}, i= 0,1,2,...,9$, then $S_{9}=\frac{x-x^{n+1}}{1-x}=\frac{1.12-(1.12)^{10} }{1-1.12}=16.5487351$
Kamal Salem
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I think you're getting downvoted because OP's question is asking for a specific "finite n" answer rather than the infinite series limit. – DotCounter Aug 13 '21 at 17:50
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I see the confusion, however, when the title calls for a "series", in my mind typically implying the limit of partial sums – sendit Aug 13 '21 at 17:51
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A sum of form $\sum_{m=1}^n r^m = r+r^2+r^3+...+r^n$ can be found in the following way:
Let $S_n = \sum_{m=1}^n r^m$ be the $n$th partial sum. Note $rS_n = r^2+r^3+...r^{n+1}$. Then $rS_n -S_n = r^{n+1}+r^n+...+r^3+r^2-(r^n+r^{n-1}+...+r^2)-r=r^{n+1}-r$. So factoring the left side yields $S_n = \frac{r^{n+1}-r}{r-1}$. So in your example we'd have $\sum_{m=1}^9 (1.12)^m = \frac{1.12^{10}-1.12}{1.12-1}=16.549$.Your best chance for calculating $\frac{1.12^{10}-1.12}{1.12-1}$ without a calculator might be to use $1.12 = 1+\frac{3}{25} = \frac{28}{25}$ and expand to get a (big) rational number.
sendit
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