Math Problem Statement

Use summation notation to write the sum: 7 + 14 + 28 + ... + 896, 4/5 + 5/6 + 6/7 + ..., -1 + 2 - 3 + 4 - 5 + 6 ... + 20, and 1 + 1/2 + 2/3 + 6/4 + 24/5 + 120/6.

Solution

Let's break down the four questions from your image and express them in summation notation as required:

12) 7+14+28++8967 + 14 + 28 + \cdots + 896

This is a geometric sequence where:

  • First term a=7a = 7
  • Common ratio r=2r = 2

The general term ana_n of a geometric sequence is given by: an=arn1=72n1a_n = a \cdot r^{n-1} = 7 \cdot 2^{n-1} The last term an=896a_n = 896, so we can write this in summation notation: n=1N72n1,where N is the term number for 896.\sum_{n=1}^{N} 7 \cdot 2^{n-1}, \quad \text{where } N \text{ is the term number for } 896.

To find NN, solve 72N1=8967 \cdot 2^{N-1} = 896: 2N1=8967=1282N1=27N1=7N=82^{N-1} = \frac{896}{7} = 128 \quad \Rightarrow \quad 2^{N-1} = 2^7 \quad \Rightarrow \quad N-1 = 7 \quad \Rightarrow \quad N = 8 Thus, the summation is: n=1872n1\sum_{n=1}^{8} 7 \cdot 2^{n-1}


13) 45+56+67+\frac{4}{5} + \frac{5}{6} + \frac{6}{7} + \cdots

This looks like a sequence of the form: n+3n+4,n=1,2,3,\frac{n+3}{n+4}, \quad n = 1, 2, 3, \dots The general term can be written as n+3n+4\frac{n+3}{n+4}, starting from n=1n = 1.

Thus, the summation notation is: n=1n+3n+4\sum_{n=1}^{\infty} \frac{n+3}{n+4} This is an infinite series, as no final term is given.


14) 1+23+45+6+20-1 + 2 - 3 + 4 - 5 + 6 - \cdots + 20

This is an alternating series where the signs alternate between positive and negative for consecutive terms.

The general term can be expressed as (1)n+1n(-1)^{n+1} \cdot n, where n=1,2,3,n = 1, 2, 3, \dots, and the last term is 2020.

Thus, the summation notation is: n=120(1)n+1n\sum_{n=1}^{20} (-1)^{n+1} \cdot n


15) 1+12+23+64+245+12061 + \frac{1}{2} + \frac{2}{3} + \frac{6}{4} + \frac{24}{5} + \frac{120}{6}

This series follows a pattern where the numerators are factorials, i.e., 1!,2!,3!,4!,5!,6!1!, 2!, 3!, 4!, 5!, 6!.

Thus, the general term can be written as: (n1)!n\frac{(n-1)!}{n} where n=1,2,3,,6n = 1, 2, 3, \dots, 6.

So the summation notation is: n=16(n1)!n\sum_{n=1}^{6} \frac{(n-1)!}{n}


Would you like more details or clarifications on any of these summations?

5 Related Questions:

  1. How do you find the sum of a geometric series?
  2. What is the formula for the sum of an infinite series?
  3. How do alternating series converge?
  4. Can factorial terms be generalized for other sequences?
  5. How would you find the sum of a finite arithmetic sequence?

Tip:

When working with series, always check if it's geometric, arithmetic, or follows a unique pattern (like factorials) to simplify finding a closed form.

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Math Problem Analysis

Mathematical Concepts

Geometric Sequences
Rational Numbers
Alternating Series
Factorials

Formulas

General term for a geometric sequence: a_n = a * r^(n-1)
Summation of alternating series: (-1)^(n+1) * n
Factorials: (n-1)!/n

Theorems

Geometric series summation
Series of rational numbers
Alternating series convergence
Factorial notation

Suitable Grade Level

Grades 9-12