Question

Write the sum using summation notation, assuming the suggested pattern continues.

-1 + 2 + 5 + 8 + ... + 44

Answer 1

I don really get how to do this

Answer 2

16
\[\sum_{n=1}^{16}3n-4\]

Answer 3

i dont think thats it but thank you anyways

Answer 4

Notice the difference from one addend to the next is 3. So the formula will be 3n + something (x). For f(1)=-1 or 3*1 + x=-1. 3+x=-1. x=-4. This makes the formula -4. We know the starting n is 1. The last addend is 44. 3n-4=44. 3n=48. n=16.

\[\sum_{n=1}^{16}3n-4\]

Answer 5

I just saw that you were asking for the sum also. To do that we separate the sigma notations.

\[\sum_{n=1}^{16}3n-4\] becomes \[3*\sum_{n=1}^{16}n -\sum_{n=1}^{?16}4\]

We use the formula n(n+1)/2 for the sum of the first n digits.

Plugging in 16 we get 136.

3*136=408

Now the second sigma is just adding 4 to itself 16 times. Or more simply put multiplying 4*16, which is 64.

Now we have 408-64=344.