Question

The sum of the first 30 terms of the sequence an = 6n + 5 is _____.

How do I create a formula to do this & what would the sum be??

Answer 1

First term plus last term, divided by two and times the number of terms:
\[S_n=\frac{ a_1+a_n }{ 2 }·n\]

Answer 2

So you want a sum like this:\[S_{30}=\sum_{i=1}^{30}(6i+5) \]By using properties of sums, you can simplify this:\[S_{30}=\sum_{i=1}^{30}6i + \sum_{i=1}^{30}5 = 6\sum_{i=1}^{30}i + \sum_{i=1}^{30}5\]Now because\[\sum_{i=1}^{30}5 = 5\times 30 = 150\]It simplifies to \[6\sum_{i=1}^{30}i + 150\]All that's left to really figure out is the sum of i from 1 to 30, which is given by the triangle area formula (see the story about Gauss):\[\sum_{i=1}^{n}i = \frac{n(n+1)}{2}\] which in this case would be \[\frac{30\times(30+1)}{2} = 465\]So your final answer would be \[465+150 = 615\]

Answer 3

thank you @CarlosGP!

Answer 4

\[a_1=11\]\[a_{30}=30 \times 6+5=185\]\[n=30\]\[S_{30}=\frac{ 11+185 }{ 2 }\times 30=2940\]

Answer 5

@alexray19 thank you so much for breaking down the steps for me! you're awesome! Where you put 30x(30+1) where did you get that?

Answer 6

@alexray19 review your rationale, it is not correct

Answer 7

@CarlosGP what is the difference between the answer you gave me & alex's answer of 615? I really can't get this question wrong :(

Answer 8

Put 2940, it is guaranteed!

Answer 9

omg thank you so much! it was right! @CarlosGP :)

Answer 10

you are welcome! happy to know youve scored

Answer 11

@alexray19 formula is right but he forgot to multiply the 465 times 6 before adding 150
6·465+150=2790+150=2940