Discrete mathematics question

Question

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

Prove the statement using mathematical induction.

Answer 1

Let $s(n)=1^2+2^2+...+n^2=\sum_{j=1}^nj^2.$. You want calculate $s(2n)-s(n)$. We know that $s(n)=\frac{n(n+1)(2n+1)}{6}$, then $s(2n)=\frac{2n(2n+1)(4n+1)}{6}$. It follows that

$$s(2n)-s(n)=\frac{2n+1}{6}(8n^2+2n-n^2-n)=\frac{2n+1}{6}(7n^2+n)=\frac{n(2n+1)(7n+1)}{6}.$$

Answer 2

We have to prove $$(n+1)^2+(n+2)^2+(n+3)^2+\ldots+(2n)^2=\frac{n(2n+1)(7n+1)}{6}$$

$\color{red}{\text{Step}\ 1}$: Setting $n=1$ in the above identity, we get $$(1+1)^2=\frac{1(2\cdot 1+1)(7\cdot 1+1)}{6}$$ $$4=\frac{3\times 8}{6}$$ $$4=4$$

$\color{red}{\text{Step} \ 2}$: Assuming the identity holds for $n=k$ then we get
$$\color{red}{(k+1)^2+(k+2)^2+(k+3)^2+\ldots+(2k)^2}=\color{blue}{\frac{k(2k+1)(7k+1)}{6}}$$

$\color{red}{\text{Step}\ 3}$: Setting $n=k+1$ on both the sides of the given identity, we get $$(k+1+1)^2+(k+1+2)^2+(k+1+3)^2+\ldots+(2k)^2+(2k+1)^2+(2(k+1))^2=\frac{(k+1)(2(k+1)+1)(7(k+1)+1)}{6}$$ $$\iff (k+2)^2+(k+3)^2+(k+4)^2+\ldots+(2k)^2+(2k+1)^2+4(k+1)^2$$ $$=\frac{(k+1)(2k+3)(7k+8)}{6}$$ $$\iff (k+1)^2+(k+2)^2+(k+3)^2+(k+4)^2+\ldots+(2k)^2+(2k+1)^2+3(k+1)^2$$ $$=\frac{(k+1)(2k+3)(7k+8)}{6}$$

$$\iff \color{red}{(k+1)^2+(k+2)^2+(k+3)^2+\ldots+(2k)^2}$$ $$=\frac{(k+1)(2k+3)(7k+8)}{6}-(2k+1)^2-3(k+1)^2$$ $$=\frac{(k+1)(2k+3)(7k+8)-6(2k+1)^2-18(k+1)^2}{6}$$ $$=\frac{(k+1)[(2k+3)(7k+8)-18(k+1)]-6(2k+1)^2]}{6}$$ $$=\frac{(k+1)[14k^2+19k+6]-6(2k+1)^2}{6}$$ $$=\frac{(k+1)[(2k+1)(7k+6)]-6(2k+1)^2}{6}$$ $$=\frac{(2k+1)[(k+1)(7k+6)-6(2k+1)]}{6}$$ $$=\frac{(2k+1)[7k^2+k)]}{6}$$ $$=\color{blue}{\frac{k(2k+1)(7k+1)}{6}}$$ Which is true from (2), hence identity holds for $n=k+1$

Hence, the given identity holds for all positive integers $n\geq 1$

Answer 3

When $n=1$, $LHS=4=RHS.$

Assume, when $n=k$, $LHS=RHS$. When $n=k+1$,

$LHS=(k+2)^2+(k+3)^2+\dots+(2k+1)^2 + (2(k+1))^2 \\=k(2k+1)(7k+1)/6-(k+1)^2+(2k+1)^2+(2(k+1))^2 \\=[k(2k+1)(7k+1)+18(k+1)^2+6(2k+1)^2]/6 \\=[(2k+1)(7k^2+13k+6)+18(k+1)^2]/6 \\=[(2k+1)(7k+6)(k+1)+18(k+1)^2]/6 \\=(k+1)(14k^2+37k+24)/6 \\=(k+1)(2(k+1)+1)(7(k+1)+1)/6=RHS.$