Question

Find\[\sum_{k=n+1}^{3n}(3k+2)\]

Answer 1

start from 1 to 3n then subtract what you get from 1 to n

Answer 2

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

Answer 3

\[ 3 \sum_{k=n+1}^{3n}k + 2\sum_{k=n+1}^{3n} 1\\
3( {3n+1 - (n+1)\over 2} ) (n+1 + 3n) + 2 (3n+1 - (n+1))\]

Answer 4

Is it
12n^2+4n

Answer 5

Total number of terms is 2n

Answer 6

after algebra, i got \(12n^2+7n\) maybe i messed up
is there a slicker way to do it?

Answer 7

maybe substitute k-n = p ....

Answer 8

what do u suggest @siddhantsharan ?

Answer 9

Are you guys sure about the answer?

Answer 10

I think its either 12n^2 +7n or 12n^2+4n

Answer 11

first one

Answer 12

satellite73 is correct