How do you find the sum of the finite geometric sequence of #Sigma (5/2)^(n-1)# from n=1 to 10?

Question

William Abdalla · Accepted Answer

#sum_(n=1)^10 (5/2)^(n-1)=3254867/512=6357.162109375#

#sum_(n=1)^10 (5/2)^(n-1)# #=sum_(m=0)^9 (5/2)^(m)# (you can still use #n# as the variable, but in effect you are letting #m=n-1#) #=((5/2)^10-1)/((5/2)-1)# (geometric series formula) #=3254867/512=6357.162109375#

From the series

#sum_(n=0)^x k^n = 1 + k + k^2 + ... + k^x#

We multiply throughout by k

#ksum_(n=0)^x k^n = k + k^2 + k^3 + ... + k^(x+1)#

Now subtract the original series from that sum

#ksum_(n=0)^x k^n - sum_(n=0)^x k^n = (k-1)sum_(n=0)^x k^n#

#= cancel(k) + cancel(k^2) + cancel(k^3) + ... + cancel(k^x) + k^(x+1) - (1 + cancel(k) + cancel(k^2) + ... + cancel(k^x)) #

#= k^(x+1) - 1#

So, dividing by (k-1), we have #sum_(n=0)^x k^n = (k^(x+1) - 1)/(k-1)#

Christopher Agresta · Answer

undefined To find the sum of the finite geometric sequence given by $ \sum_{n=1}^{10} \left(\frac{5}{2}\right)^{n-1} $, you can use the formula for the sum of a finite geometric series:

$$ S_n = \frac{a(1 - r^n)}{1 - r} $$

where:
- $ S_n $ is the sum of the series,
- $ a $ is the first term of the series,
- $ r $ is the common ratio of the series, and
- $ n $ is the number of terms in the series.

In this sequence, $ a = 1 $ (the first term), $ r = \frac{5}{2} $ (the common ratio), and $ n = 10 $ (the number of terms). Plug these values into the formula:

$$ S_{10} = \frac{1(1 - \left(\frac{5}{2}\right)^{10})}{1 - \frac{5}{2}} $$

$$ S_{10} = \frac{1 - \left(\frac{5}{2}\right)^{10}}{1 - \frac{5}{2}} $$

$$ S_{10} = \frac{1 - \frac{9765625}{1024}}{\frac{-3}{2}} $$

$$ S_{10} = \frac{1 - 9537.1953125}{-\frac{3}{2}} $$

$$ S_{10} = \frac{-9536.1953125}{-\frac{3}{2}} $$

$$ S_{10} = \frac{9536.1953125}{\frac{3}{2}} $$

$$ S_{10} = 9536.1953125 \times \frac{2}{3} $$

$$ S_{10} = 6357.46354167 $$

Therefore, the sum of the given finite geometric sequence is approximately $ 6357.46 $.