Edexcel A-Level Mathematics AS Fp1 8 Proof Questions

For $n=1, \quad \sum_{r=1}^{1} r^{3}=1$ and $\quad \frac{1}{4}\left(1^{2}\right)(1+1)^{2}=\frac{1}{4} \times 1 \times 4=1$

So true for n=1
(Assume the result is true for n=k, so $\sum_{r=1}^{k} r^{3}=\frac{1}{4} k^{2}(k+1)^{2}$ )
$$\text { Then } \quad \sum_{r=1}^{k+1} r^{3}=\frac{1}{4} k^{2}(k+1)^{2}+(k+1)^{3}$$$=\frac{1}{4}(k+1)^{2}\left[k^{2}+4(k+1)\right]=\frac{1}{4}(k+1)^{2}\left[k^{2}+4 k+4\right]$$=\frac{1}{4}(k+1)^{2}(k+2)^{2}$
Hence result is true for n=k+1. As true for n=1 and have shown if true
for n=k then it is true for n=k+1, so it is true for all $n \in \mathbb{N}$ by induction.
(5)

$n=1, \mathrm{LHS}=1^{2}=1, \mathrm{RHS}=\frac{1}{6} \cdot 2 \cdot 3=1$

Shows both LHS =1 and RHS =1
Accept LHS = 1 but must see at
least $\frac{1}{6} \cdot 2 \cdot 3=1$ for RHS.
Assume true for n=k

When n=k+1
$$\sum_{r=1}^{k+1} r^{2}=\frac{k}{6}(k+1)(2 k+1)+(k+1)^{2}$$

Adds $(k+1)^{2}$ to result for n=k
$=\frac{(k+1)}{6}(k(2 k+1)+6(k+1))$

Attempt to factorise by $\frac{(k+1)}{6}$$$=\frac{(k+1)}{6}(k+2)(2 k+3)$$$$\left.=\frac{(k+1)}{6}((k+1)+1)\right)((2(k+1)+1)$$

Either factorised form.
True for n=1.

If true for n=k then true for n=k+1 therefore true for all n.

Complete proof with no errors and
these 4 statements seen anywhere.
Depends on both M's and the A, but
may be scored if the B is lost as long
as some indication of true for n=1 is
given.

A1cso

(5)