2014年10月22日 · Let us find a formula for the nth partial sum of a geometric series. S_n=a+ar+ar^2+cdots+ar^{n-1} by multiplying by r, Rightarrow rS_n=ar+ar^2+cdots+ar^{n-1}+ar^n by subtracting rS_n from S_n, Rightarrow (1-r)S_n=a-ar^n=a(1-r^n) (Notice that all intermediate terms are cancelled out.) by dividing by (1-r), Rightarrow S_n={a(1-r^n)}/{1-r} I …

The geometric series is a series in which there is a constant ratio between consecutive terms, that is, a_n=a_{n-1}*r. This series converges if -1<r<1.

2017年8月21日 · Consider the series: #1/2+1/6+1/12+1/20+...+1/(n(n+1))+...# This is not a geometric series, but we can find its sum as follows:

2016年6月21日 · How do you find the sum of the following infinite geometric series, if it exists. 2 + 1.5 +... How do you find the sum of the first 5 terms of the geometric series: 4+ 16 + 64…? How do you find S20 for the geometric series 4 + 12 + 36 + 108 + …?

2014年9月24日 · How do you find the sum of the following infinite geometric series, if it exists. 2 + 1.5 +... How do you find the sum of the first 5 terms of the geometric series: 4+ 16 + 64…? How do you find S20 for the geometric series 4 + 12 + 36 + 108 + …?

2015年7月9日 · a+ar+ar^2+\\cdots+ar^n=\\frac{a(r^{n+1}-1)}{r-1} if r!=1 and a+a+a+\\cdots+a=(n+1)a when there are n+1 terms (and r=1). Consider a finite geometric series with n+1 terms a+ar+ar^2+\\cdots+ar^n. Call this sum S_{n}. Note that rS_{n}=ar+ar^2+ar^3+\\cdots+ar^{n+1} so that rS_{n}-S_{n}=ar^{n+1}-a. Solving this equation …

2014年9月11日 · The sum of a geometric series can be found by sum_{n=0}^infty ar^n=a/{1-r} if |r|<1.

2015年12月9日 · 18 An infinite geometric series of the form sum_(n=1)^ooar^(n-1) converges if and only if |r|<1, where r is the common ration between terms and given by r=x_(n+1)/(x_n). In this case it converges to the value a/(1-r) where a is …

2015年9月2日 · How do you find the sum of the following infinite geometric series, if it exists. 2 + 1.5 +... How do you find the sum of the first 5 terms of the geometric series: 4+ 16 + 64…? How do you find S20 for the geometric series 4 + 12 + 36 + 108 + …?

2016年3月15日 · 2186 Given that first term a_1=2 Common ratio q=a_2/a_1=6/2=3 General expression for nth term is a_n=a_1*q^(n-1) To ascertain what is number of the last term, insert in the above equation 1458=2*3^(n-1) => 3^(n-1) = 1458/2=729 Writing 729 as power of 3 we obtain 3^(n-1) = 3^6, comparing the exponents n-1=6 or n=7 Now sum of n terms is given by the …