∑7n from n = 1 to n = 100 is a shorthand method of writing :-
7x1 + 7x2 + 7x3 +...........+ 7x 99 + 7x100
This can be written as 7(1 + 2 + 3 +........+ 99 + 100)
The sum (S) of a range of numbers from 1 to n is given by the formula S = 1/2 n(n + 1)
And substituting the relevant numbers gives, S = 1/2 x 100 x 101 = 5050
But this now needs multiplying by 7 to give the sum of the multiples = 7 x 5050 = 35350.
There are infinitely many multiples of 21 and it is impossible to list them all other than as 21*n where n is an integer.
There are infinitely many multiples. They are all of the form 41*k where k is any integer.There are infinitely many multiples. They are all of the form 41*k where k is any integer.There are infinitely many multiples. They are all of the form 41*k where k is any integer.There are infinitely many multiples. They are all of the form 41*k where k is any integer.
Yes. By definition a multiple of 8 is any number that can be expressed as 8*n, where n is an integer. But 8n=4*(2*n), and 2*n is an integer, when n is an integer. Because 8n equals four times an integer, 8n is a multiple of 4.
I don't know about "magic" in this context. Multiples are the number 9, multiplied by an integer, for example: 0 x 9 = 0 1 x 9 = 9 2 x 9 = 18 etc. If you just keep adding 9 at a time, you will get all the multiples.
They are all the numbers in the set {41472*k, where k is an integer}.
Any multiple of 700. 700, 1400, 2100 and so on...
18 and all the multiples of 18.
Z3 = {3n | n is an integer}
There are infinitely many multiples of 21 and it is impossible to list them all other than as 21*n where n is an integer.
There are infinitely many multiples of 5 so it is not possible to list them. They are all of the form 5*k where k is an integer.
Least common multiple is 88. Other multiples are 176, 264, 352, etc. and all integer multiples of 88.
There are infinitely many multiples of 10 so it is far from clear what you mean by "all 5" multiples. Any integer which ends in a 0 is a multiple of 10.
All multiples of 32 are also multiples of 8.
There are infinitely many multiples. They are all of the form 41*k where k is any integer.There are infinitely many multiples. They are all of the form 41*k where k is any integer.There are infinitely many multiples. They are all of the form 41*k where k is any integer.There are infinitely many multiples. They are all of the form 41*k where k is any integer.
All numbers of the form 552*k where k is an integer.
They are all integers of the form 40*k where k is an integer.
Yes. By definition a multiple of 8 is any number that can be expressed as 8*n, where n is an integer. But 8n=4*(2*n), and 2*n is an integer, when n is an integer. Because 8n equals four times an integer, 8n is a multiple of 4.