Hi
In like this question in
I stop in same point and and I can't complete next part of question please help
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) $\displaystyle \frac{k(k+1)(2k+1)}{6}+(k+1)^2$
$\displaystyle (k+1)\left[\frac{k(2k+1)}{6}+(k+1)\right]$
$\displaystyle (k+1)\frac{(2k^2+7k+6)}{6}$
$\displaystyle \frac{(k+1)(k+2)(2k+3)}{6}$
$\displaystyle \frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}$
Therefore the statement is true for P(k+1).
2) $\displaystyle \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$
$\displaystyle (k+1)(k+2)\left[\frac{k}{3}+1\right]$
$\displaystyle (k+1)(k+2)\left(\frac{k+3}{3}\right)$
$\displaystyle \frac{(k+1)[(k+1)+1][(k+1)+2]}{3}$
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.
$\displaystyle P(n) : (xy)^n=x^{n}y^{n}$
P(1) : Left hand side = $\displaystyle (xy)^1=xy$
Right hand side = $\displaystyle x^{1}y^{1}=xy$
Therefore the statement is true for P(1).
Suppose that the statement is true for P(k). Then,
$\displaystyle P(k) : (xy)^k=x^{k}y^{k}$
Therefore, $\displaystyle (xy)^{k}(xy)=x^{k}y^{k}(xy)$
$\displaystyle (xy)^{k+1}=x^{k+1}y^{k+1}$
Therefore the statement is true for P(1).
Hope this will help you.