Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers n :

$1+4+7+...+(3n-2)=\frac{1}{2}n(3n-1)$

Short answer

The statement is true for all$n\in N$

Step-by-step solution

. Given Information

We are given a statement :

$1+4+7+...+(3n-2)=\frac{1}{2}n(3n-1)$

. Checking for n=1

$P\left(1\right)=1(L.H.S)$

$P\left(1\right)=\frac{1(3-1)}{2}=1(R.H.S)$

Therefore the statement is true for n=1

. Checking for n=k

Let P(n) is true for n=k so we need to prove that P(k+1) is true :$P\left(k\right)=1+4+7+...+(3k-2)=\frac{1}{2}k(3k-1)$ -equation(i)

So, the next term will be :

$1+4+7+...+(3k-2)+(3k+1)=\frac{(k+1)(3k+2)}{2}$

Now we need to modify the L.H.S so that it is equal to the R.H.S, we can substitute the value of equation one in this :

$\frac{k(3k-1)+2(3k+1)}{2}=\frac{3k^2-k+6k+2}{2}$

$\frac{3k^2+5k+2}{2}=\frac{(k+1)(3k+2)}{2}$

So, R.H.S = L.H.S which means P(n) is true for n=k+1

Hence , the statement is true for all$n\in N$