Question

Find Σ where the top limit is 18 and the bottom is i=7 and its multiplied bt (11i+3)


1,686

1,653

1,620

1,545.5

please I am so confused

Answer 1

heres a picture

Answer 2

You mean:\[\sum_{i=1}^{18}\left(11i+3\right)\]? You can separate this:\[11\underbrace{\sum_1^{18}i}_{\frac{18\cdot19}2}+\underbrace{\sum_1^{18}3}_{3\cdot18}\]

Answer 3

It's a summation problem, so you have to add as many terms together until you get to your upper limit. So, to start, the first term is 11*7+3=80. Replace the 7 with 8, then 9, and so on until 18, adding each term together.

Answer 4

Notably\[\sum_{i=1}^ni=\frac{n(n+1)}2\]which motivates the first part of my solution.

Answer 5

yeah badrefrences perfect

Answer 6

please help I have no idea

Answer 7

I gave you the answer already. You just need a calculator. What's the problem?

Answer 8

whats n?

Answer 9

Okay, let me rewrite this.

Answer 10

a. 1,686

b. 1,653

c. 1,620

d. 1,545.5

Answer 11

\[\sum_{i=1}^{18}(11i+3)=11\sum_{i=1}^{18}i+\sum_{i=1}^{18}3\\\sum_{i=1}^{18}i=\frac{18\cdot19}2\\\sum_{i=1}^{18}3=3\cdot18\]Please don't tell me you're too lazy to finish this.

Answer 12

(11*7+3)+(11*8+3)+(11*9+3)...(11*18+3)

Answer 13

ooh okay thank you and I'm not lazy I haven't learned geometric so its kinda hard but thank you

Answer 14

answer is 1686,
\[\sum_{i=7}^{18}(11i+3)\]=\[\sum_{i=7}^{18}11i\]+3*(18-7+1)=\[11\sum_{i=7}^{18}i\]+36=11*(7+18)*0.5+36=1686