1. Let the roots of $x^2-x-1=0$ be $\alpha,\beta$.

By Vieta's formulas, $\alpha+\beta=1, \qquad \alpha\beta=-1.$

We are given $p_k=\alpha^k+\beta^k.$ We need to check which statement is not true.

2. Derive a recurrence for $p_k$.

Since each root satisfies $r^2=r+1,$ for $r=\alpha,\beta$, multiplying by $r^{k-2}$ gives $r^k=r^{k-1}+r^{k-2}.$ So, $\alpha^k=\alpha^{k-1}+\alpha^{k-2},$ $\beta^k=\beta^{k-1}+\beta^{k-2}.$ Adding, $p_k=p_{k-1}+p_{k-2}.$

Initial values: $p_1=\alpha+\beta=1,$ $p_2=\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=1^2-2(-1)=3.$

Now compute successively: $p_3=p_2+p_1=3+1=4,$ $p_4=p_3+p_2=4+3=7,$ $p_5=p_4+p_3=7+4=11.$

3. Check each option.

Option A

$(p_1+p_2+p_3+p_4+p_5)=1+3+4+7+11=26.$ So A is true.

Option B

$p_5=11.$ So B is true.

Option C

$p_5-p_4=11-7=4=p_3.$ So C is true.

Option D

$p_2\cdot p_3=3\cdot 4=12 \neq 11=p_5.$ So D is false.

4. Therefore, the statement which is not true is $\boxed{D}.$