Mathematical Induction getting the right side

Question

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 $$\frac{3(5^{k+2} −1)}{4}.$$

Answer 1

(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

Answer 2

$$\frac{k(k+1)(k+2)}{6} + (k+1)^2$$

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: $$\frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}$$ $$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6} $$ Factoring out a $(k+1)$ from the numerator: $$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$$ Expanding the expression inside the parentheses: $$=\frac{(k+1)(2k^2+k+6k+6)}{6}$$ $$ = \frac{(k+1)(2k^2+7k+6)}{6}$$ Now we can factor $2k^2+7k+6$ into $(k+2)(2k+3)$ and we obtain: =$$ \frac{(k+1)(k+2)(2k+3)}{6}$$

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In the second part of your question the two equations are not equal. If you want me to explain how I could simplify the expression into something a little more clean, feel free to comment!

Answer 3

Instead of expressing the RHS as $$\frac{(k+1)(k+2)(2k+3)}{6}$$ I would be inclined to express it as $$\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$ Similarly, for the LHS, I would prefer to express it in terms of $k+1$ instead of $k$. $$\frac{k(k+1)(2k+1)}{6}+(k+1)^2 = \frac{((k+1)-1)(k+1)(2(k+1)-1)}{6}+(k+1)^2$$ Now of that looks a bit messy, try making a substitution, for example $m=k+1$. Then the LHS would become $$\frac{(m-1)m(2m-1)}{6}+m^2$$ and the RHS becomes $$\color{blue}{\frac{m(m+1)(2m+1)}{6}}$$ Continuing with the LHS we have $$\begin{array}{lll} \displaystyle\frac{(m-1)m(2m-1)}{6}+m^2 &=& \displaystyle\frac{(m-1)m(2m-1)}{6}+\frac{6m^2}{6}\\ &=& \displaystyle\frac{(m-1)m(2m-1)}{6}+\frac{6m^2}{6}\\ &=&\displaystyle\frac{m}{6}((m-1)(2m-1)+6m)\\ &=&\displaystyle\frac{m}{6}(2m^2-3m+1+6m)\\ &=&\displaystyle\frac{m}{6}(2m^2+3m+1)\\ &=&\displaystyle\frac{m}{6}(2m+1)(m+1)\\ &=&\displaystyle\color{blue}{\frac{m(m+1)(2m+1)}{6}}\\ \end{array}$$