grenivkah3z

2022-07-04

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
$\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1{\right)}^{2}$
And I'm trying to make it equal to
$\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}$
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.
$\frac{3\left({5}^{k+1}-1\right)}{4}+3·{5}^{k+1}$
to
$\frac{3\left({5}^{k+2}-1\right)}{4}.$

Oliver Shepherd

Expert

(1)
$\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1{\right)}^{2}=\frac{k\left(k+1\right)\left(2k+1\right)+6\left(k+1{\right)}^{2}}{6}=\left(k+1\right)\frac{k\left(2k+1\right)+6\left(k+1\right)}{6}=\left(k+1\right)\frac{k\left(2k+1\right)+6k+6}{6}=\left(k+1\right)\frac{2{k}^{2}+7k+6}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}$
(2) Is your second statement even true? for $k=1$ I get: $34=93$
Let ${5}^{k}=x$
$\frac{3\left(5x-1\right)}{4}+3x+1=\frac{15x-3+12x+4}{4}=\frac{27x+1}{4}$
which doesn't seem to be equal to your given RHS

Gretchen Schwartz

Expert

$\frac{k\left(k+1\right)\left(k+2\right)}{6}+\left(k+1{\right)}^{2}$
Let us get the part of this expression on the right side to have a common denominator with the fraction, let us use 6. We now have:
$\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\frac{6\left(k+1{\right)}^{2}}{6}$
$=\frac{k\left(k+1\right)\left(2k+1\right)+6\left(k+1{\right)}^{2}}{6}$
Factoring out a $\left(k+1\right)$ from the numerator:
$=\frac{\left(k+1\right)\left(k\left(2k+1\right)+6\left(k+1\right)\right)}{6}$
Expanding the expression inside the parentheses:
$=\frac{\left(k+1\right)\left(2{k}^{2}+k+6k+6\right)}{6}$
$=\frac{\left(k+1\right)\left(2{k}^{2}+7k+6\right)}{6}$
Now we can factor $2{k}^{2}+7k+6$ into $\left(k+2\right)\left(2k+3\right)$ and we obtain: =
$\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}$

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