Question

$1.3+3.5+5.7+...........+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$ is true for

• A. Only natural number n $\ge$ 4
• B. Only natural numbers 3 $\le$ n $\le$ 10
• C. All natural numbers n
• D. None

Answer 1

The correct option is C All natural numbers n

Le P(n) be the given statement
i.e. P(n) : 1.3 + 3.5 + 5.7 + ........... + (2n -1)(2n+1)
$=\frac{n(4n^2+6n-1)}{3}$
Putting $n=1,L.H.S.=1.3=3$ and $RHS=\frac{1({4.1}^2+6.1-1)}{3}$
$=\frac{4+6-1}{3}=\frac{9}{3}=3$
LHS $=$ RHS
$\therefore$ P(n) is true for n $=$ 1
Assume that P(n) is true for n $=$ k
i.e., p(k) is true
i.e., P(k) : 1.3 + 3.5 + 5.7 + ............ + (2k-1)(2k+1)
$=\frac{k(4k^2+6k-1)}{3}$
Last term $=$ (2k -1)(2k +1)
Replacing k by (k+1), we get
$\left[2\right(k+1)-1]\left[2\right(k+1)+1]=(2k+1)(2k+3)$
$\therefore$ Adding (2k+1)(2k+3) on both sides.
$\therefore LHS=1.3+3.5+5.7+...........+(2k-1)(2k+1)+(2k+1)(2k+3)$
$R.H.S=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3)$
$=\frac{(4k^3+6k^2-k)+3(2k+1)(2k+3)}{3}$
$=\frac{4k^3+18k^2+23+9}{3}$
$=\frac{(k+1)(4k^2+14k+9)}{3}$
$=\frac{(k+1)\left(4\right(k+1{)}^2+6(k+1)-1)}{3}$ ............ (iii)
Thus, P(n) is true for n $=$ k + 1
$\therefore$ P(k+1) is true whenever P(k) is true
Hence, by P(n) is true for all n $\in$ N