evaluate the R(1(3))
Answers
You can take the sum of a finite number of terms of a geometric sequence. And, for reasons you'll study in calculus, you can take the sum of an infinite geometric sequence, but only in the special circumstance that the common ratio r is between –1 and 1; that is, you have to have | r | < 1.
For a geometric sequence with first term a1 = a and common ratio r, the sum of the first n terms is given by:
\displaystyle{ \sum_{i=1}^n \, a_i = a\left(\dfrac{1 - r^n}{1 - r}\right) }i=1∑nai=a(1−r1−rn)
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Note: Your book may have a slightly different form of the partial-sum formula above. For instance, the "a" may be multiplied through the numerator, the factors in the fraction might be reversed, or the summation may start at i = 0 and have a power of n + 1 on the numerator. All of these forms are equivalent, and the formulation above may be derived from polynomial long division.
In the special case that | r | < 1, the infinite sum exists and has the following value:
\displaystyle{ \sum_{i=1}^{\infty}\,a_i = \dfrac{a}{1 - r} }i=1∑∞ai=1−ra
Evaluate the following:
\mathbf{\color{green}{\displaystyle{ \sum_{\mathit{i}=1}^{20}\,3(-2)^{\mathit{i}} }}}i=1∑203(−2)i
The first few terms are –6, 12, –24:
a1 = 3(–2)1 = (3)(–2) = –6
a2 = 3(–2)2 = (3)(4) = 12
a3 = 3(–2)3 = (3)(–8) = –24
So this is a geometric series with common ratio r = –2. (I can also tell that this must be a geometric series because of the form given for each term: as the index increases, each term will be multiplied by an additional factor of –2.)
The first term of the sequence is a = –6. Plugging into the summation formula, I get:
(-6)\left(\dfrac{1 - (-2)^{20}}{1 - (-2)}\right)(−6)(1−(−2)1−(−2)20)
= (-6)\left(\dfrac{1 - 1,048,576}{1 + 2}\right)=(−6)(1+21−1,048,576)
= (-6)\left(\dfrac{-1,048,575}{3}\right)=(−6)(3−1,048,575)
= (-2)(-1,048,575)=(−2)(−1,048,575)
= 2,097,150=2,097,150
So the value of the summation is:
2,097,150
Evaluate S10 for 250, 100, 40, 16,....
The notation "S10" means that I need to find the sum of the first ten terms. The first term is a = 250. Dividing pairs of terms, I get:
100 \div 250 = \frac{2}{5}100÷250=52
40 \div 100 = \frac{2}{5}40÷100=52
...and so forth, so the terms being added form a geometric sequence with common ratio r = \frac{2}{5}r=52.
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Unlike the formula for the n-th partial sum of an arithmetic series, I don't need the value of the last term when finding the n-th partial sum of a geometric series. So I have everything I need to proceed. When I plug in the values of the first term and the common ratio, the summation formula gives me:
\mathrm{S}_{10} = 250\left(\dfrac{1 - \left(\frac{2}{5}\right)^{10}}{1 - \frac{2}{5}}\right)S10=250(1−521−(52)10)
= 250\left(\dfrac{1 - \frac{1,024}{9,765,625}}{\frac{3}{5}}\right)=250(531−9,765,6251,024)
= 250\left(\frac{9,764,601}{9,765,625}\right)\left(\frac{5}{3}\right)=250(9,765,6259,764,601)(35)
= 250\left(\frac{3,254,867}{1,953,125}\right)=250(1,953,1253,254,867)
= \small{\dfrac{6,509,734}{15,625}}=15,6256,509,734
I will not "simplify" this to get the decimal form, because that would almost-certainly be counted as a "wrong" answer. Instead, my answer is:
\color{purple}{ \textbf{S}_{\mathbf{10}} = \mathbf{\small{ \dfrac{6,509,734}{15,625} }}}S10=15,6256,509,734
Note: If you try to do the above computations in your calculator, it may very well return the decimal approximation of 416.62297... instead of the fractional (and exact) answer.