I need help with proofs using mathematical induction: $2+7+12+17+...+(5n-3)=(\frac{n}{2})(5n-1)$

Question

I need help with this problem: $2+7+12+17+...+(5n-3)=(\frac{n}{2})(5n-1)$

Answer 1

You're looking at $$\sum_{k=1}^n (5k-3)$$

This is $$5\sum_{k=1}^n k-\sum_{k=1}^n 3=5\sum_{k=1}^nk -3n$$

Do you know what $$\sum_{k=1}^n k $$ is?

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The claim is true for $n=1$, since $$2=\frac 1 2 (5-1)$$

Assume the claim is true for $n$, and look at $n+1$. This is $$\tag 1 2+7+\cdots+5n-3+5(n+1)-3$$

The induction hypothesis is that the first $n$ terms sum up to $$\frac{n(5n-1)}2$$ thus $(1)$ is $$\frac{n(5n-1)}2+5(n+1)-3$$

Can you show this is $$=\frac{(n+1)(5(n+1)-1)}2\text{ ? }$$

Answer 2

$2+7+12+17+...+(5n-3)=(\frac{n}{2})(5n-1)$

This equality holds true for n=1,2,3....... $\ $so we suppose it will hold true for n=k

$2+7+12+17+...+(5k-3)=(\dfrac{k}{2})(5k-1)$

then we will check it for $n=k+1 $ $$2+7+12+17+...+(5k-3)+(5k+2)=(\frac{k+1}{2})(5k+4)$$ $$(\frac{k}{2})(5k-1)+(5k+2)=(\frac{k+1}{2})(5k+4)$$ now we try to prove LHS=RHS

$$\frac{(5k^2-k)+(10k+4)}{2}$$ $$\frac{(5k^2+9k+4)}{2}$$ $$\frac{(5k^2+5k+4k+4)}{2}$$ $$\frac{5k(k+1)+4(k+1)}{2}$$ $$\frac{(k+1)(5k+4)}{2}$$

hence proved..