Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction

Question

My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$

My workings

1. LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
2. $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
3. LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$

Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?

Answer 1

This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $\;k=n\;$ we get

$$(n+1)^2+(n+2)^2+\ldots+(2n)^2=\frac{n(2n+1)(7n+1)}6\;\;(**)$$

then for $\;k=n+1\;$ we get

$$(n+1+1)^2+(n+3)^2+\ldots+(2n)^2+(2n+1)^2+\overbrace{(2n+2)^2}^{=(2(n+1))^2}=\frac{(n+1)(2n+3)(7n+8)}6$$

Now, we prove the last one assuming the first one. Developing left side:

$$(n+2)^2+\ldots+(2n)^2+(2n+1)^2+(2n+2)^2=(n+1)^2+\ldots+(2n+2)^2-(n+1)^2=$$

$$=(n+1)^2+\ldots+(2n)^2+(2n+1)^2+(2n+2)^2-(n+1)^2\stackrel{\text{Ind. Hyp.}}=$$

$$=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+4(n+1)^2-(n+1)^2=$$

$$=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2=\frac{(2n+1)\left[n(7n+1)+12n+6\right]+18(n+1)^2}6=$$

$$=\frac{(2n+1)(7n^2+13n+6)+18n^2+36n+18}6=\frac{(n+1)(2n+3)(7n+8)}6$$

The last equality you can prove it opening parentheses and equating coefficients, or showing the left side has as roots $\;-1,\,-3/2,\,-8/7\;$

Alternative way to prove:

$$(n+1)^2+(n+2)^2+\ldots+(n+n)^2=$$

$$=n^2+\ldots+n^2+2n+4n+\ldots+2nn+1+2^2+\ldots+n^2=$$

$$n\cdot n^2+2(1+2+\ldots+n)n+\frac{n(n+1)(2n+1)}6=$$

$$=n^3+n^2(n+1)+\frac{n(n+1)(2n+1)}6=\ldots$$

Answer 2

Observe that:

$$(k+2)^2+\cdots+(2k+2)^2=$$$$\left[(k+1)^2+(k+2)^2+\cdots+(2k)^2\right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$

where $f(k)$ denotes the RHS of 2).

If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.

Answer 3

Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = \frac{(k+1)(2k+3)(7k+8)}{6}$$. We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = \frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$ $$ = \frac{(k+1)(2k+3)(7k+8)}{6}$$

Hope it helps

Answer 4

Assuming the step for $n=k$ $\:$For

$n=k+1$ $(k+2)^2+\cdots+(2k+2)^2=\dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=\dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=\dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=\dfrac{(k+1)(2k+3)(7k+8)}{6}$

Answer 5

$(k+2)^2+\dots+(2k)^2+(2k+1)^2+(2k+2)^2=\frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=\frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=\frac{(k+1)(2k+3)(7k+8)}6$.