<s>[INST] Mathematical Induction getting the right side So I 've been doing Mathematical Inductions but I seem to have a issue in simplify and getting the right side.
So I have this on the L.H.S
$$\frac{k(k + 1)(2k +1)}{6} + (k + 1)^2 $$
And I'm trying to make it equal to 
$$\frac{(k+1)(k+2)(2k+3)}{6}$$
I've check the book for help but for this it seems to be skipping some steps.
And I might as well post this as well which I had the same issue as above.
$$\frac{3(5^{k+1} −1)}{4}+ 3·5^{k+1}$$
to
$$\frac{3(5^{k+2} −1)}{4}.$$
 [/INST] (1)$\frac{k(k+1)(2k+1)}{6} + (k+1)^2 =\frac{k(k+1)(2k+1) + 6(k+1)^2}{6} =
(k+1)\frac{k(2k+1)+6(k+1)}{6} = (k+1)\frac{k(2k+1)+6k+6}{6} = (k+1)\frac{2k^2 + 7k + 6}{6} = \frac{(k+1)(k+2)(2k+3)}{6} $
(2)  Is your second statement even true? for $k=1$ I get: $34 = 93$...
Let $5^k = x$ 
$\frac{3(5x-1)}{4} + 3x + 1 = \frac{15x - 3 + 12x + 4}{4} = \frac{27x+1}{4}$
which doesn't seem to be equal to your given RHS
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