How can I make huge numbers with only tiny numbers? I have to make 55 with 1,1,1,2,2

Question

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Using 1,1,1,2,2, these five numbers, I have to make 55.

I think I can make from Fibonacci's function, but I can't define this with these five numbers. (I think I can't use functions for my teacher's decete answer).
So I think I can use sigma, and sum 1 to 10 also same as 55, but I don't know how to do it. Please help me.

You can use only these five numbers, 1,1,1,2,2. If you square, you also use 2, I mean that squared numbers are also included in these five numbers. You can use all math operations, factorial, square or even floor function.

Answer 1

1. If you want to use sigma, you have to reserve a one for the bottom, and so you have 1,1,2,2 to make 10

Past this point is spoilers if you still want to work it out yourself. But it is interesting and fun

——————————————————————————-

Hint: $5 \times 2 =10$ so if you can make a five with 1, 2, 1

———- First Method (Sigma) ————-

And to make ten you can do $((2+1)! -1) \times 2 = 10$

And so then move onto the sigma function $\sum_{x=1}^{10}x = 55$

------------------Second Method (Double Factorial)------------------------------

2. If you can use double factorial then you can do this method:

double factorial is when you wright for example '$6!!$' which is equal to the product of all numbers of the same parity of $6$ so $6!! = 6 \times 4 \times 2 = 48$

So here is the method

$(2 + 1) = 3$

$(2 + 1)! = 6$

$((2 + 1)!)!! = 48$

$((2 + 1)!)!! + (2 + 1)! = 54$

$((2 + 1)!)!! + (2 + 1)! + 1= 55$

---------------Third Method (Floor)-----------------------

This method uses the floor function

$(1 + 2)! = 6$

$((1 + 2)!)! = 720$

$\sqrt{((1 + 2)!)!} = 26.8328$

$\sqrt{((1 + 2)!)!} \times 2 = 53.67$

$\lfloor\sqrt{((1 + 2)!)!} \times 2\rfloor = 53$

$\lfloor\sqrt{((1 + 2)!)!} \times 2\rfloor + 1 + 1 = 55$

———————————-Method Four (Without Factorial)——————————————

This method only uses sine, and floor

$sin(1) = 0.01745$

$sin(1)^{1-2} = 57.297$

$\lfloor sin(1)^{1-2}\rfloor = 57$

$\lfloor sin(1)^{1-2}\rfloor - 2 \times 1= 55$

Answer 2

Maybe simply concatenate them:

112/2-1=55

Edit: Sorry I didn't see that this answer is already in the comments. So let me provide 1 more answer, using Fibonacci:

$F_{11-1}=F_{10}=55$

So $F_{11-1}+2-2=55$ or $F_{11-1}\times2\div2=55$