I've tried a lot of things but failed to do it, I've calculated the result inside the square root which is $7027801$ using substitution and factoring but $\sqrt{7027801}$ isn't possible to simplify.
$\endgroup$ 14 Answers
$\begingroup$I used the following remarkable identity: $$(a+b)^2 = a^2 + 2ab+b^2.$$
So $1+50\cdot 51 \cdot 52 \cdot 53=1 + 50\cdot53 \cdot (50+1) \cdot (53-1) = 1+ (50\cdot 53)^2 + 2 \cdot 50\cdot53 = (1+50\cdot53)^2$.
And hence the answer is $50\cdot 53 +1 = 2651$.
$\endgroup$ 3 $\begingroup$Ah, this is my favorite mental math trick to amaze my friends sometimes. If you are incliner to fill out the details, then check that:
$$1+x(x+1)(x+2)(x+3)=(x^2+3x+1)^2$$
$\endgroup$ $\begingroup$Let $x=51$.
We have $\sqrt{1 + (x-1)\cdot x\cdot(x+1)\cdot(x+2)}$.
Simplifying, multiplying out, we get $\sqrt{x^4+2x^3-x^2-2x+1}$.
This is also $\sqrt{(x^2+x-1)^2}$.
Note that $x^2 + x - 1 = 2651$ since $x=51$.
$\endgroup$ 1 $\begingroup$Including $50\cdot51\cdot 52\cdot 53$ inside the square root suggests that you should choose a value for $x$. Suppose you choose the highest value and set $x=53$. Then
\begin{align}1+x(x-1)(x-2)(x-3)&=1+x(x-3)\cdot(x-1)(x-2)\\& =1+(x^2-3x)(x^2-3x+2) \end{align}
Now let $y=x^2-3x$. The above equation becomes
$$1+y(y+2)=y^2+2y+1=(y+1)^2$$
therefore
$$\sqrt{(y+1)^2}=y+1=x^2-3x+1={53}^2-3(53)+1=2651$$
$\endgroup$ 2
