Question

In Problems 26–28, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers.

$1^2+4^2+7^2+.......{(3n-2)}^2=\frac{1}{2}n(6n^2-3n-1)$

Short answer

Given statement is true for all natural numbers.

Step-by-step solution

Step 1. Given information. 

Statement:

$1^2+4^2+7^2+.......{(3n-2)}^2=\frac{1}{2}n(6n^2-3n-1)$

Step 2. Show that statement is true for n=1.

$$\begin{gathered} L.H.S.=1^2 \\ =1 \end{gathered}$$

$$\begin{gathered} R.H.S.=\frac{1}{2}.1(6.1^2-3.1-1) \\ =\frac{1}{2}(6-3-1) \\ =\frac{1}{2}(6-4) \\ =\frac{2}{2} \\ =1 \end{gathered}$$

Since$L.H.S.=R.H.S.$

Hence, statement is true for 1.

Step 3. Assume that statement is true for n=k.

$1^2+4^2+7^2+.......{(3k-2)}^2=\frac{1}{2}k(6k^2-3k-1)$

Step 4. Show that the statement is true for n=k+1.

$$\begin{gathered} L.H.S.=1^2+4^2+7^2+.......{(3k-2)}^2+(3{(k+1)-2)}^2 \\ =\frac{1}{2}k(6k^2-3k-1)+{(3k+3-2)}^2 \\ =\frac{1}{2}k(6k^2-3k-1)+{(3k+1)}^2 \\ =\frac{1}{2}(6k^3-3k^2-k)+(9k^2+1+18k) \\ =\frac{(6k^3-3k^2-k)+2(9k^2+1+18k)}{2} \\ =\frac{(6k^3-3k^2-k+18k^2+2+36k)}{2} \\ =\frac{6k^3+15k^2+35k+2}{2} \end{gathered}$$Use result of step 3.

$$\begin{gathered} R.H.S.=\frac{1}{2}(k+1)(6{(k+1)}^2-3(k+1)-1) \\ =\frac{1}{2}(k+1)\left(6\right(k^2+2k+1)-3k-3-1) \\ =\frac{1}{2}(k+1)\left(6k^2+12k+6-3k-4\right) \\ =\frac{1}{2}(k+1)(6k^2+9k+2) \\ =\frac{6k^3+15k^2+35k+2}{2} \end{gathered}$$

Since $L.H.S.=R.H.S.$for $n=k+1$

Hence the statement is true for all natural number