Question

Please solve it's urgent​

Answer 1

p(m + 1) = (m + 1)²

Given :-

$\sf{p(m) = 1 + 3 + 5 + .........+ (2m - 1) = m^{2}}$

To Find :-

$\sf{Sum \: of \: (m + 1) \: terms \: of \: this \: series -}$

$\: \: \: \:$$\sf{p(m + 1) = ?}$

Usable Formula :-

This is a sequence of sum of the consecutive odd numbers .

We know :-

$\sf{1 + 3 + 5 + ..............+ (2m - 1) }$

$\sf{Sum \: of \: first \: m \: odd \: number = m^{2}}$

$\sf{Sum \: of \: first \: n \: terms \: of \: a \: consecutive \: series -}$

$\: \: \: \:$$\sf{Sn =\dfrac{n}{2}[a + An]}$

As Confirmation :-

$\: \: \:$$\sf{Sm =\dfrac{m}{2}[1 + (2m - 1)]}$

$\: \: \:$$\sf{Sm =\dfrac{m}{2}(2m)}$

$\: \: \:$$\sf{Sm = m\times m = m^{2}}$

Solution :-

$\sf{A(m + 1) = 2(m + 1) - 1 = 2m + 2 - 1 = 2m + 1}$

$\sf{p(n + 1) = 1 + 3 + 5 + .......+ (2m + 1)}$

Here ,

$\: \: \: \:$$\sf{first \: term \: (a) = 1}$

$\sf{(m + 1)th \: term = (2m + 1)}$

Then ,

$\sf{P(m + 1) =\dfrac{(m + 1)}{2}[1 + (2m + 1)]}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf{=\dfrac{(m + 1)}{2}(2m + 2)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf{=\dfrac{(m + 1)}{2}\times 2(m + 1)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf{= (m + 1)^{2}}$