Question

Use sigma notation to represent the sum of the first seven terms of the following sequence -4,-6,-8.....

Answer 1

Answer:Answer:

$\sum\left {{7} \atop {1}} \right -n(3+n)$

Step-by-step explanation:

Given the sequence -4,-6,-8..., in order to get sigma notation to represent the sum of the first seven terms of the sequence, we need to first calculate the sum of the first seven terms of the sequence as shown;

The sum of an arithmetic series is expressed as $S_n = \frac{n}{2}[2a+(n-1)d]$

n is the number of terms

a is the first term of the sequence

d is the common difference

Given parameters

n = 7, a = -4 and d = -6-(-4) = -8-(-6) = -2

Required

Sum of the first seven terms of the sequence

$S_7 = \frac{7}{2}[2(-4)+(7-1)(-2)]\\\\S_7 = \frac{7}{2}[-8+(6)(-2)]\\\\S_7 = \frac{7}{2}[-8-12]\\\\\\S_7 = \frac{7}{2} * -20\\\\S_7 = -70$

The sum of the nth term of the sequence will be;

$S_n = \frac{n}{2}[2(-4)+(n-1)(-2)]\\\\S_n = \frac{n}{2}[-8+(-2n+2)]\\\\S_n = \frac{n}{2}[-6-2n]\\\\S_n = \frac{-6n}{2} - \frac{2n^2}{2}\\S_n = -3n-n^2\\\\S_n = -n(3+n)$

The sigma notation will be expressed as $\sum\left {{7} \atop {1}} \right -n(3+n)$. The limit ranges from 1 to 7 since we are to  find  the sum of the first seven terms of the series.