# Math Help - I want complete this answers >> help me

1. ## I want complete this answers >> help me

Hi

In like this question in

here I attached three question and I want complete the answers with explain

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2. Originally Posted by r-soy
Hi

In like this question in

here I attached three question and I want complete the answers with explain

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Dear r-soy,

In each of the problems you could simplify your last expressions further,

1) $\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$(k+1)\left[\frac{k(2k+1)}{6}+(k+1)\right]$

$(k+1)\frac{(2k^2+7k+6)}{6}$

$\frac{(k+1)(k+2)(2k+3)}{6}$

$\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}$

Therefore the statement is true for P(k+1).

2) $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

$(k+1)(k+2)\left[\frac{k}{3}+1\right]$

$(k+1)(k+2)\left(\frac{k+3}{3}\right)$

$\frac{(k+1)[(k+1)+1][(k+1)+2]}{3}$

Therefore the statement is true for P(k+1).

3) I think there is a bit of confusion in this problem. So I have done it from the beginning.

$P(n) : (xy)^n=x^{n}y^{n}$

P(1) : Left hand side = $(xy)^1=xy$

Right hand side = $x^{1}y^{1}=xy$

Therefore the statement is true for P(1).

Suppose that the statement is true for P(k). Then,

$P(k) : (xy)^k=x^{k}y^{k}$

Therefore, $(xy)^{k}(xy)=x^{k}y^{k}(xy)$

$(xy)^{k+1}=x^{k+1}y^{k+1}$

Therefore the statement is true for P(1).