Question

How do I find the sum from k=l to 30 of (4k-3)? I know I can do the long way and plug in each number from 1-30, but is there an easier way like a power sum equation I'm not seeing?

Answer 1

\[\sum_{k=1}^{30}(4k-3)\]

Answer 2

nice question

Answer 3

ur really cute girl dear

Answer 4

thanks, do you have a solution for my question?

Answer 5

\[\sum_{k=1}^{30}(4k-3)=4\sum_{k=1}^{30}k-\sum_{k=1}^{30}3 \]
Recall that: \[ \sum_{k=1}^nk=\frac{k(k+1)}{2}\]
And if \(c\) is a constant, than \[ \sum_{k=1}^nc=nc\]

So:
\[ 4\sum_{k=1}^{30}k-\sum_{k=1}^{30}3 =4\left( \frac{30(30+1)}{2}\right)-3(30)\]

Answer 6

Sorry typo: 2nd line should read:
\[ \sum_{k=1}^nk=\frac{n(n+1)}{2}\]

Answer 7

thank you so much!