# Prove using mathematical induction that for all n≥1, 1+4+7+...+(3n-2)= (n (3n-1)) / 2 Hello!

We have to prove that

`sum_(k=1)^n a_k=sum_(k=1)^n (3k-2) = (n(3n-1))/2` for all n>=1.

First, establish the base of induction for n=1:

`sum_(k=1)^n (3k-2)=3*1-2=1` and `(1*(3*1-1))/2=1.`

These numbers are equal, so the base is established. Now, assume that for some n the statement is true and consider the statement for...

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Hello!

We have to prove that

`sum_(k=1)^n a_k=sum_(k=1)^n (3k-2) = (n(3n-1))/2` for all n>=1.

First, establish the base of induction for n=1:

`sum_(k=1)^n (3k-2)=3*1-2=1` and `(1*(3*1-1))/2=1.`

These numbers are equal, so the base is established. Now, assume that for some n the statement is true and consider the statement for n+1:

`sum_(k=1)^(n+1) (3k-2) = ((n+1)(3(n+1)-1))/2.`

Whether it is true? The sum at the left is eqial to

`sum_(k=1)^(n) (3k-2)+a_(n+1), = sum_(k=1)^(n) (3k-2)+(3(n+1)-2)`

which is equal to `(n(3n-1))/2+(3(n+1)-2)` by the induction assumption, and this is the same as `(3n^2-n+6n+2)/2=(3n^2+5n+2)/2.`

The expression at the right is `((n+1)(3n+2))/2=(3n^2+3n+2n+2)/2,` which is the same.

This way, the induction step is also proved and the initial statement is proved also.

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