Question

Prove the following by PMI for all $n$ belong to $N$
$1\times 3+3\times 5+5\times 7.........+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$

Answer 1

When $n=1$ we have the end term of the series as $(2\ast 1-1)(2\ast 1+1)=1\ast 3=3$
Putting $n=1$ in the R.H.S of the given equation we have
$\frac{1(4\ast 1^2+6\ast 1-1)}{3}=\frac{1(4+6-1)}{3}=3$
Therefore the equation is valid for $n=1$
Let the expression be valid for any value $n=k$ where 'k' belongs to N.
So $1.3+3.5+.....+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3}$ holds true. ---------eqn(1)
Now we have to prove that the equation is valid for n=k+1.
i.e. $1.3+3.5+....+\left[2\right(k+1)-1]\left[2\right(k+1)+1]=\frac{(k+1)\left(4\right(k+1{)}^2+6(k+1)-1)}{3}$----eqn(2)
Now L.H.S of equation 2 can be written as
$1.3+3.5+....+(2k-1)(2k+1)+\left[2\right(k+1)-1]\left[2\right(k+1)+1]$-----------expression(1)
Putting the value of the R.H.S of equation 1 in expression 1 we have
$\frac{k(4k^2+6k-1)}{3}+(2k+2-1)(2k+2+1)$

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

or $\frac{(4k^3+6k^2-k)}{3}+\frac{3(4k^2+6k+2k+3)}{3}$

or $\frac{(4k^3+6k^2-k)}{3}+\frac{(12k^2+24k+9)}{3}$

or $\frac{(4k^3+6k^2-k+12k^2+24k+9)}{3}$

or $\frac{(4k^3+18k^2+23k+9)}{3}$

or $\frac{(4k^3+4k^2+14k^2+14k+9k+9)}{3}$

or $\frac{4k^2(k+1)+14k(k+1)+9(k+1)}{3}$

or $\frac{(k+1)(4k^2+14k+9)}{3}$

or $\frac{(k+1)(4k^2+8k+6k+4+6-1)}{3}$

or $\frac{(k+1)4k^2+8k+4+6k+6-1}{3}$

or $\frac{(k+1)\left(4\right(k^2+2k+1)+6(k+1)-1)}{3}$

or $\frac{(k+1)4(k+1{)}^2+6(k+1)-1}{3}$-----expression(2)

But expression 2 is nothing but the R.H.S. of the equation 2.
Therefore by mathematical induction we have proved that the said equation holds true for every value of 'n' where 'n' belongs to N.