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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?

S17
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  • You should make sure you're clear about what is actually true and what you're trying to prove. $8+11+14+\dots+(3(k+1)+5)= \frac12(k+1)(3(k+1)+13)$ is what you're trying to prove; the subsequent statements are manipulations of the left-hand side. – Patrick Stevens Mar 26 '17 at 13:22
  • Sorry, proving it's true for all n∈N. – S17 Mar 26 '17 at 13:24
  • Induction proofs have two things 1) Is the base case true? in this case for $n=1$ is this true? 2) Assume this is true for the $n^{th}$ case. Will that imply $(n+1)^{th}$ case is then true? – Sentinel135 Mar 26 '17 at 13:33
  • BTW we can treat your equations as a sumation namely $\sum^n_{k=1} 3k+5 = \frac{n}{2}(3n+13)$ – Sentinel135 Mar 26 '17 at 13:34

2 Answers2

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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=$$\frac{1}{2}n(3n+13)+3n+8=\frac{3n^2+13n+6n+16}{2}=\frac{3n^2+19n+16}{2}=\frac{3n^2+3n+16n+16}{2}=\frac{(n+1)(3n+16)}{2}=\frac{(n+1)(3(n+1)+13)}{2}$$

LM2357
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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 $\forall n \in \mathbb{Z^+}$

mrnovice
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