Find the values of ## n\geq 1 ## for which ## 1+2+3+\dotsb +n ##.

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In summary: Thus, in summary, we have shown that for all values of ##n\geq 1##, except for 1 and 3, the sum ## 1!+2!+3!+\dotsb +n! ## is not a perfect square. Therefore, the only values for which the sum is a perfect square are ##n=1, 3##.
  • #1
Math100
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Homework Statement
Find the values of ## n\geq 1 ## for which ## 1!+2!+3!+\dotsb +n! ## is a perfect square.
Relevant Equations
None.
Let ## f(n)=1!+2!+3!+\dotsb +n! ## for ## n\in\mathbb{N} ##.
Then every ## k! ## is even except ## 1! ##.
This means ## f(n) ## is odd for ## n>1 ##.
Since each ## k! ## is divisible by ## 5 ## for ## k\geq 5 ##,
it follows that ## f(4)\pmod {5}\equiv (1+2+6+24)\pmod {5}\equiv 3\pmod {5} ##.
Thus, ## f(n)\equiv 3\pmod {2} ## and ## f(n)\equiv 3\pmod {5} ## for ## n>3 ##, which implies that ## f(n)\equiv 3\pmod {10} ##.
Note that ## f(1)=1^{2}, f(2)=3 ## and ## f(3)=3^{2} ##, where ## f(2)=3 ## isn't a perfect square.
Therefore, the values of ## n\geq 1 ## for which ## 1!+2!+3!+\dotsb +n! ## is a perfect square are ## n=1, 3 ##.
 
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  • #2
This is just a checking of your result.

for n ##\geq## 8 f(n)##\equiv 0##(mod 9)
So f(n) is written as ##f(n)=3^2 A##
for n ##\geq## 9 f(n)##\equiv 3##(mod 10)
So the first place of A is 7. Any square number does not have 7 in its first place. So f(n) for n ##\geq## 9 is not square number.

for n ##\geq## 1 f(n)##\equiv 1##(mod 2)
for n ##\geq## 2 f(n)##\equiv 0##(mod 3)
for n ##\geq## 3 f(n)##\equiv 1##(mod 4)
for n ##\geq## 4 f(n)##\equiv 3##(mod 5)
for n ##\geq## 5 f(n)##\equiv 3##(mod 6)
for n ##\geq## 6 f(n)##\equiv 5##(mod 7)
for n ##\geq## 7 f(n)##\equiv 1##(mod 8)
for n ##\geq## 8 f(n)##\equiv 0##(mod 9)
for n ##\geq## 9 f(n)##\equiv 3##(mod 10)
 
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  • #3
Math100 said:
Therefore, the values of ## n\geq 1 ## for which ## 1!+2!+3!+\dotsb +n! ## is a perfect square are ## n=1, 3 ##.
You have not proven that. You have only shown that 1 and 3 are such values and 2 is not. You need to prove there are no values higher than 3. You could try using your mod results to do that.
 
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  • #4
andrewkirk said:
You have not proven that. You have only shown that 1 and 3 are such values and 2 is not. You need to prove there are no values higher than 3. You could try using your mod results to do that.
Yes, true but he has prove that ##f(n)=3 \pmod {10}## for ##n>3## and it is kind of trivial to show that ##a^2 \pmod {10}\neq 3## for any natural number ##a##, so ##f(n)\neq a^2## for any natural number ##a##..
 
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  • #5
Math100 said:
Homework Statement:: Find the values of ## n\geq 1 ## for which ## 1!+2!+3!+\dotsb +n! ## is a perfect square.
Relevant Equations:: None.

Let ## f(n)=1!+2!+3!+\dotsb +n! ## for ## n\in\mathbb{N} ##.
I would deal with the small cases by hand:

##f(1)=1 ## and ##f(3)=9## are perfect squares, ##f(2)=3## and ##f(4)=33## are not.
Math100 said:
Then every ## k! ## is even except ## 1! ##.
This means ## f(n) ## is odd for ## n>1 ##.
Since each ## k! ## is divisible by ## 5 ## for ## k\geq 5 ##,
it follows that ## f(4)\pmod {5}\equiv (1+2+6+24)\pmod {5}\equiv 3\pmod {5} ##.
A typo? It follows ##f(n) \equiv f(4) \equiv 3 \pmod 5.## Now conclude with @Delta2 's proposal in post #4:

##a^2 \equiv c \in \{0,1,4\} \pmod 5## so ##f(n)\neq a^2## for any integer ##a.##

Math100 said:
Thus, ## f(n)\equiv 3\pmod {2} ## and ## f(n)\equiv 3\pmod {5} ## for ## n>3 ##, which implies that ## f(n)\equiv 3\pmod {10} ##.
Note that ## f(1)=1^{2}, f(2)=3 ## and ## f(3)=3^{2} ##, where ## f(2)=3 ## isn't a perfect square.
Therefore, the values of ## n\geq 1 ## for which ## 1!+2!+3!+\dotsb +n! ## is a perfect square are ## n=1, 3 ##.
 
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FAQ: Find the values of ## n\geq 1 ## for which ## 1+2+3+\dotsb +n ##.

1. What is the formula for finding the sum of consecutive numbers from 1 to n?

The formula for finding the sum of consecutive numbers from 1 to n is (n*(n+1))/2.

2. How do you prove that the formula for finding the sum of consecutive numbers is correct?

The formula can be proven using mathematical induction, where the base case n=1 is true and the formula holds for all subsequent values of n.

3. What is the significance of finding the sum of consecutive numbers?

Finding the sum of consecutive numbers has many practical applications, such as calculating the total cost of a series of items or determining the total amount of time spent on a task.

4. Is there a limit to the value of n that can be used in the formula?

No, the formula can be used for any value of n greater than or equal to 1.

5. Can the formula be applied to find the sum of consecutive odd or even numbers?

Yes, the formula can be applied to find the sum of consecutive odd or even numbers by adjusting the value of n to only include the desired numbers.

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