Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. Picture attached.

Answer 1

True

1) There are three steps, to using mathematical induction.

2) The first one is to check for n=1

`\begin{gathered} \left(3n-2\right)^2=(n\left(6n^2-3n-1\right))/(2) \\ \\ \left(3\cdot \:1-2\right)^2=(1\cdot \left(6\cdot \:1^2-3\cdot \:1-1\right))/(2) \\ \\ (3-2)^2=(1(6-3-1))/(2) \\ \\ 1^2=(2)/(2) \\ \\ 1=1\:\:True \end{gathered}`

3) Secondly, let's prove it for n=k:

`\sum _(k=1)^k\left(3k-2\right)^2=(k\left(6k^2-3k-1\right))/(2)`

4) And finally, for n=k+1

`\begin{gathered} \sum _(n=1)^n\left(3n-2\right)^2=(n\left(6n^2-3n-1\right))/(2) \\ \\ \sum _(\left(k+1\right)=1)^(k+1)\left(3\left(k+1\right)-2\right)^2=(\left(k+1\right)\left(6\left(k+1\right)^2-3\left(k+1\right)-1\right))/(2) \\ \\ \left(3\left(k+1\right)-2\right)^2+(k\left(6k^2-3k-1\right))/(2)=(\left(k+1\right)\left(6\left(k+1\right)^2-3\left(k+1\right)-1\right))/(2) \\ \\ 2\left(3\left(k+1\right)-2\right)^2+k\left(6k^2-3k-1\right)=\left(k+1\right)\left(6\left(k+1\right)^2-3\left(k+1\right)-1\right) \\ \\ 6k^3+15k^2+11k+2=6k^3+15k^2+11k+2 \\ \\ 6k^3+15k^2+11k+2-2=6k^3+15k^2+11k+2-2 \\ \\ 6k^3+15k^2+11k=6k^3+15k^2+11k \\ \\ 6k^3+15k^2+11k-\left(6k^3+15k^2+11k\right)=6k^3+15k^2+11k-\left(6k^3+15k^2+11k\right) \\ \\ 0=0 \\ \\ True. \end{gathered}`

As all of those 3 steps were true. Then we can tell, that this is true.