#1**0 **

Although the notation is sloppy, I will assume based on former questions of the same that "∑n=15[5(−2)^n−1]" translates mathematically to \(\sum_{n=1}^5[5(-2)^{n-1}]\). Please correct me, though, if this in incorrect.

Anyway, there is neat formula that will deal with question very quickly. It is the following:

\(S_n=a_1\left(\frac{1-r^n}{1-r}\right),r\neq1\)

S_{n }is the sum of the geometric series

a_{1 }is the first term in the series

r is the common ratio

n is the number of terms in the finite geometric series

Let's use this formula to our advantage! However, there are two variables that we do not know: a_{1 }and n. Let's start with a_{1}.The first term in the series is when n=1, so let's determine what that is.

\(a_1=5(-2)^{1-1}\) | A lot of simplification will occur here--especially since the exponent simplifies to zero. |

\(a_1=5\) | |

The number of terms also must be calculated:

\(n=5-1+1\) | To find the number of terms, subtract the maximum n, 5, from the minimum, one. Add one to make up for the missing one. |

\(n=5\) | We can do the formula! |

\(a_1=5,n=5,r=-2;\\ S_n=a_1\left(\frac{1-r^n}{1-r}\right)\) | Do the substitution! |

\(S_n=5\left(\frac{1-(-2)^5}{1-(-2)}\right)\) | Let's simplify the numerator and denominator. |

\(S_n=5\left(\frac{1-(-32)}{1+2}\right)\) | |

\(S_n=5\left(\frac{33}{3}\right)\) | |

\(S_n=5*11=55\) | Just like that, we are done! |

TheXSquaredFactor May 17, 2018