Can you find the square root?

Let $n=666$ , then the given expression reduces to: $\sqrt{n(n+1)(n+2)(n+3)+1}$

= $\sqrt{n^{4}+6n^{3}+11n^{2}+6n+1}$

= $\sqrt{n^{2}(n^{2}+6n+11+\frac{6}{n}+\frac{1}{n^{2}})}$

= $n\sqrt{n^{2}+\frac{1}{n^{2}}+6(n+\frac{1}{n})+11}$

= $n\sqrt{(n+\frac{1}{n})^{2}-2+6(n+\frac{1}{n})+11}$

= $n\sqrt{(n+\frac{1}{n})^{2}+6(n+\frac{1}{n})+9}$

= $n(n+\frac{1}{n}+3)$

= $n^{2}+3n+1$

Putting value of n=666, We get the answer 445555

Let $n = 666$ the equation then rewrites to

$\sqrt {n\cdot(n+1)\cdot(n+2)\cdot(n+3) + 1}$

Multiplying out the brackets gives us

$\sqrt{(n^2 + n)\cdot(n + 2)\cdot(n+3) + 1}$

$\sqrt{(n^3 + 3n^2 + 2n)\cdot(n + 3)+1}$

$\sqrt{n^4 + 6n^3 + 11n^2 + 6n+1}$

So what's the square root of that?

Let's try looking at it in a different way, how about we put together the two outermost brackets and the two innermost brackets.

$\sqrt{(n+1)\cdot(n+2)-1} \cdot \sqrt{n\cdot(n+3)+1}$

Note -

$(n+1)\cdot(n+2)- 1=n\cdot(n+3) + 1$

$n^2 + 3n +2 - 1 = n^2 + 3n + 1$

So the square root of the equation is equal to both

$(n+1)\cdot(n+2) - 1$

$and$

$n\cdot(n+3) + 1$

Inputting the value of $n$ into either one gives us

$(666)^2 + 3(666) + 1 = 445,555$

Shortcut method

When a, b, c, d are consecutive numbers then

(a × b × c × d + 1)^1/2 then answer is (a × d + 1) or (b × c – 1)

= (666 × 669 + 1) or = (667 × 668 – 1)

Squaring both sides of $\sqrt{666 \times 667 \times 668 \times 669 + 1} = x$ and subtracting 1 from both sides: $666 \times 667 \times 668 \times 669 = (x+ 1)(x-1)$

LHS is even, hence RHS = product of two successive even numbers.

Taking the product of middle two and extreme two will give even numbers close to each other. Products (N)(N+3) and (N+1)(N+2) indeed turn out to be $N^2+3N$ and $N^2+3N+2$ Which are successive evens!

Hence $x-1 = (N)(N+3)$ and since N is 666, $x = 666 \times 669 + 1 = \boxed{445555}$

If the product of a, b, c, and d; four consecutive natural numbers plus 1. It would always be perfect square. square root trick would work for it as average(a d,b c).or (a d+b c)/2. so it would be (666 669+667 668)/2=(445554+445556)/2=445555.

A property can be derived by using the Arithmetic progression....

now, root( 1x2x3x4+1) = 5 again root(2x3x4x5+1) = 11... so as u progress the answer is a ROOT.... so get the difference between these terms is in the following 6, 8, 10, 12...... and on.....

So the common difference is 2.... now get the sum till the 665 terms with the first term as 6..... so,it boils down to 665x670 and add 5 so the answer is 445555....... so, here a property is derived ..... take minus 1 term of the first term and multiply with the Added 1 to the last term and add 5....... this can be done for all the squences......

Multiplication of four consecutive numbers plus 1 always is a perfect square. And the square root of the ans will same as the mean of the multiplication of 1st and last term and middle two terms.