Answer:
See below.
Step-by-step explanation:
I will assume that 3n is the last term.
First let n = k, then:
Sum ( k terms) = 7k^2 + 3k
Now, the sum of k+1 terms = 7k^2 + 3k + (k+1) th term
= 7k^2 + 3k + 14(k + 1) - 4
= 7k^2 + 17k + 10
Now 7(k + 1)^2 = 7k^2 +14 k + 7 so
7k^2 + 17k + 10
= 7(k + 1)^2 + 3k + 3
= 7(k + 1)^2 + 3(k + 1)
Which is the formula for the Sum of k terms with the k replaced by k + 1.
Therefore we can say if the sum formula is true for k terms then it is also true for (k + 1) terms.
But the formula is true for 1 term because 7(1)^2 + 3(1) = 10 .
So it must also be true for all subsequent( 2,3 etc) terms.
This completes the proof.
To start this, try to place it all together.
(x)+(x+2)+(x+4)=36
Then you start to combine so that it looks like this:
3x+6=36
I believe that is one of the choices. Do you need more explanation?
