Question

A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely.

Sn: 1 + 4 + 7 + . . . + (3n - 2) = n(3n - 1)/2

Answer 1

If we have

`\displaystyle S_n: \sum_(i=1)^n (3i-2)=(n(3n-1))/(2)`

in order to write
`S_k` and
`S_(k+1)`, we simply have to substitute "n" with "k" and "k+1", respectively.

So, we have

`\displaystyle S_k: \sum_(i=1)^k (3i-2)=(k(3k-1))/(2)`

`\displaystyle S_(k+1): \sum_(i=1)^(k+1) (3i-2)=((k+1)(3(k+1)-1))/(2)=((k+1)(3k+3-1))/(2)=((k+1)(3k+2))/(2)=(3k^2+5k+2)/(2)`