Prove that the following statement is true; $$8+11+14+...+(3n+5)= \frac12n(3n+13)$$So far I have P(1) is true. Assuming that P(k) is true; $$8+11+14+...+(3k+5)= \frac12k(3k+13)$$ Then I need to deduce that P(k+1) is true so, $$8+11+14+...+(3(k+1)+5)= \frac12(k+1)(3(k+1)+13)$$ $$8+11+14+...+(3(k+1)+5)$$ $$(8+11+14+...+(3k+5))+(3(k+1)+5)$$ $$\frac12k(3k+13)+(3(k+1)+5), by P(k)$$ $$\frac12k(3k+13)+(3k+8)$$
Not really sure where to go from here? Or have I gone wrong somewhere?
On
It seems all correct. Going from your last line:
$$LHS = \frac12k(3k+13)+(3k+8)$$
$$=\frac{3k^2+13k+6k+16}{2} = \frac{3k^2+19k+16}{2}$$
Then notice that $k=-1$ is a root of that quadratic
$$LHS = \frac{(k+1)(3k+16)}{2} = \frac{k+1}{2}(3(k+1)+13)$$
as required, hence true $\forall n \in \mathbb{Z^+}$
So you were all correct until now
This is true for base case $1$ so..
Let us assume true for $n$.
$\sum^n_{k=1}3k+5=\frac12n(3n+13)$
Adding $(3(n+1)+5)$ to both sides
R.H.S=$$\frac{1}{2}n(3n+13)+3n+8=\frac{3n^2+13n+6n+16}{2}=\frac{3n^2+19n+16}{2}=\frac{3n^2+3n+16n+16}{2}=\frac{(n+1)(3n+16)}{2}=\frac{(n+1)(3(n+1)+13)}{2}$$