Proving $\sum_{r=n+1}^{2n}(3r+4) =\frac n2(9n+11)$ (mathematical Induction)

Question

Hope someone can help me understand what to apply here, currently stuck. I know you are meant to do summation. However I try to find to solve problems like this and I'm not finding much resources.

$$\sum_{r=n+1}^{2n}(3r+4) =\frac n2(9n+11)$$

Answer 1

You need to find:

$$\sum_{n=k+1}^{2k} 3n+4 $$ The number of terms in this sum is $2k-(k+1)+1 = k$. So the sum is $$ 3(k+1 + k+2 + k+3 + \cdots + 2k) + 4(k) \\ = 3(\underbrace{k + k + k \cdots + k)}_{k \ times} + 3(1+2+\cdots + k) +4k $$ Consider the sequence $1+2+\cdots + k$.

Now note that $k+1=(k-1)+2=(k-2)+3=\cdots$, ie the sum of of the $n$th element from the left hand of the sequence and the $n$th element from the right hand is equal for all possible values of $n$. In total, this value is added $\frac{k}{2}$ times, since pairing is done in groups of $2$.

So the sum becomes:

$$3k^2 + \frac{3(k+1)(k)}{2} + 4k$$ $$\frac{k}{2}\cdot (6k+3k+3+8)$$ $$\frac{k}{2}\cdot (9k+11)$$