Question

In Problem, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n$.

$1·2+3·4+5.6+...+(2n-1)\left(2n\right)=\frac{1}{3}n(n+1)\left(\right(4n-1)$

Short answer

The statement is shown.

Step-by-step solution

Step 1. Given information

The statement is$1·2+3·4+5.6+...+(2n-1)\left(2n\right)=\frac{1}{3}n(n+1)\left(\right(4n-1)$.

Step 2. Show the given statement.

Put $n=1$in the statement $1·2+3·4+5.6+...+(2n-1)\left(2n\right)=\frac{1}{3}n(n+1)\left(\right(4n-1)$

width="177">1.2=13,1.(1+1)(4.1-1)

So, the statement is true for $n=1$.

Next, we assume that the above statement is true for some $k$. so that

width="359">1.2+3.4+5.6+⋯+(2k-1)(2k)=13k(k+1)(4k-1)

Now we consider the sum of first $k+1$terms.

$\begin{array}{rcl}1.2+3.4+5.6+\cdots +(2k-1)\left(2k\right)+\left(2\right(k+1)-1)\left(2\right(k+1\left)\right)& =& \frac{1}{3}k(k+1)(4k-1)+(2k+2-1)\left(2\right(k+1\left)\right)\\ & =& \frac{(k+1)}{3}\left(k\right(4k+3)+2(4k+3\left)\right)\\ & =& \frac{(k+1)}{3}\left(4k^2+3k+8k+6\right)\\ & =& \frac{1}{3}(k+1)\left(\right(k+1)+1)\left(4\right(k+1)-1)\end{array}$

As a result, statement is true for all natural numbers.