The expression 1!+2!+3!+4!++n! (Where n is a natural number) is not a / an: a) composite number. b) odd number. c) perfect square. d) multiple of 3

c) perfect square

Step-by-step explanation:

Lets take one example:

n = 4

1 + 2! + 3! + 4! = 33

33 is odd, multiple of 3 and also composite. But, not a perfect square

Good morning ☕️

The expression 1!+2!+3!+4!++n! (Where n is a natural number) is not a / an:

Step-by-step explanation:

1!+2!+3!+4!++n! Is a composite number

proof: 1!+2!=3 and 3!+4!++n! has 3 as a factor

then 1!+2!+3!+4!++n! has 3 as a factor

1!+2!+3!+4!++n! is an odd number

proof: 2!+3!+4!++n! has 2 as a factor then it’s an even number

then if we add 1! , we get an odd number 1!+2!+3!+4!++n!

1!+2!+3!+4!++n! is a multiple of 3

proof : 1!+2!=3 and 3!+4!++n! has 3 as a factor

then 1!+2!+3!+4!++n! has 3 as a factor therefore it’s a multiple of 3.

1!+2!+3!+4!++n! is not a perfect square

proof: a perfect square number must end with one of these digits 0,1,4,5,6 and 9

in other words it will never end with 3

reason :

0^2=0

1^2=1

2^2=4

3^2=9

4^2=16

5^2=25

6^2=36

7^2=49

8^2=64

9^2=81

in addition, the factorial of any number from 5 to n end with zero

cause it has 2 and 5 as factors at the same time

examples:

5!=120

6!=720

7!=5 040

.

.

.

then the sum 5!++n! Will end with 0

also, 1!+2!+3!+4!=33

conclusion:

1!+2!+3!+4!++n!  = (1!+2!+3!+4!) +(5! ++n!)

                             = 33 + ( a large number which end with 0)

                             = a large number that ends with 3

we conclude that :1!+2!+3!+4!++n! is not a perfect square

:)

where's the diagram?

step-by-step explanation: