Is That Too Much to Calculate?(2)

Notice that $a_n=(n + 1)^2 + (n + 2)^2 + ((n + 1)(n + 2))^2$

$=((n+1)(n+2))^2+2n^2+6n+5$

$=((n+1)(n+2))^2+2(n^2+3n+2)+1$

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

So $a_{124}=(125\times126+1)^2, r=15751$

Recall that $\begin{aligned} I & = A^2+ B^2+ C^2\\& = \,(A +B +C)^2-2\,(AB+BC+CA) \end{aligned}$ which follows as $\begin{aligned} a_X& \overset{\mathrm{fr. above }}{ =} \,(X+1)^2 +\,(X+2)^2 +\,(X+1)^2\,(X+2)^2\\& \overset{\mathrm{simplify}}{= }\left(X^2+5X+5\right)^2 -4\,(X+1)\,(X+2)^2 \\ & \overset{\mathrm{simplify}}{=}\,(X+2)^4 -2\,(X+1)\,(X+2)^2 +\,(X+1)^2 \\& = \,(\,(X+2)^2-(X+1))^2 \\r^2 & =\,(X^2+3X+3)^2 \end{aligned}$ Thus, $a_{124}\,(r) =X(X+3)+3 = 124 \times 127 +3 =\boxed{15751}$

$a_n = x^2+y^2+z^2$

$a_{124} = 125^2+126^2+15750^2={\boxed{248094001}}$

$r^2 = 248094001$

$r = {\sqrt{248094001}} = {\boxed{15751}}$

hence the answer is $\color{#3D99F6}{\boxed{15751}}$

Note: we have expressed $x,y,z$ in form of $a_n$ as below.

• $x = n+1$

• $y = n+2$

• $z = (x×y)$

In my honest opinion, there should be written one more example - to the $a_{4}$ . This is important because there is more than one rule of understanding the rule, which I had at the beginning. I tried to make it in the way of thinking that to make the last basis of exponentiation in $a_{n}$ I should multiply the basis of exponentiation in the middle by $2$ , and then add it to the last basis of exponentiation in $a_{n-1}$ . I hope you will understand what I mean ;). Btw. the answer for my rule is $r=\surd 62757901$ :P

$a_n = (n + 1)^2 + (n + 2)^2 + ((n + 1)(n + 2))^2$ $\implies a_{124} = (125)^2 + (126)^2 + (125 \times 126)^2 = r^2$

$\implies r = \sqrt{218094001}$

$\implies r = 15751$