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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.

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    $\begingroup$ Generalize: try and factor $(a-1)\cdot a \cdot(a+1)\cdot(a+2)$ $\endgroup$
    – Integrand
    Aug 5 '20 at 16:15
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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 $50\cdot 53 +1 = 2651$.

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  • $\begingroup$ Did it take you more than a minute to factor that? Cause frankly I wouldn't ever have made that connection. $\endgroup$
    – Anthony
    Aug 5 '20 at 18:52
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    $\begingroup$ Actually it took me roughly a minute. When you want to get rid of the square root you want to make appear a square. Here 1 was alone so I knew this was going to be something like $(1+a)^2$. Then I played with the numbers as 51 is close to 50 and 52 to 53. $\endgroup$
    – Axel
    Aug 5 '20 at 21:25
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    $\begingroup$ We could have done it differently: $1+50 \cdot 51 \cdot 52 \cdot 53 = 1+(51-1)\cdot 51 \cdot 52 \cdot (52+1) = 1+ (51\cdot52)^2-2\cdot 51 \cdot 52 = (51\cdot 52-1)^2 = (2652-1)^2 $ $\endgroup$
    – Axel
    Aug 5 '20 at 21:28
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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:

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

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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 $x=51$.

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  • $\begingroup$ I had tried this but haven't even comprehended the fact that $x^4+2x^3-x^2-2x+1=(x^2+x-1)^2$ ! $\endgroup$
    – Anthony
    Aug 5 '20 at 16:35
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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

$$\sqrt{(y+1)^2}=y+1=x^2-3x+1={53}^2-3(53)+1=2651$$

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  • $\begingroup$ That factorization skills though! $\endgroup$
    – Anthony
    Aug 5 '20 at 16:37
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    $\begingroup$ @Antonio: I've explained my approach by the substitution $y=x^2-3x$. The same logic will work for $1+x(x+1)(x+2)(x+3)$ with the substitution $y=x^2+3x$. $\endgroup$
    – Axion004
    Aug 6 '20 at 16:14

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