- #1
heaven eye
- 30
- 0
a series such as :-
∑ from n = 1 ,until m for n
can be found with the law of the sum of arithmetical series:-
[m(2a+(m-1)d)]/2
where :-
m= the number of terms
a= the first term in the series
d= the basic arithmetical
in the pervious example :-
a=1 , d=1 , m=m
when we solve the last information in the law we find :-
∑ from n = 1 ,until m for n = m(m+1)/2
for example :-
1+2+3+4+...+18+19+20 = 20(20+1)/2 = 210
my question is :-
what about the series :-
∑ from n = 1 ,until m for n^2 ?
it isn't an artithmetical or even a geometrical series then :-
what kind of series is it ?
and how could they find that :-
∑ from n = 1 ,until m for n^2 = [m(m+1)(2m+1)]/6 ?
for example :-
(1^2)+(2^2)+(3^2)+(4^2)+(5^2)= 5(5+1)(2*5+1)/6 = 55
and thank you
∑ from n = 1 ,until m for n
can be found with the law of the sum of arithmetical series:-
[m(2a+(m-1)d)]/2
where :-
m= the number of terms
a= the first term in the series
d= the basic arithmetical
in the pervious example :-
a=1 , d=1 , m=m
when we solve the last information in the law we find :-
∑ from n = 1 ,until m for n = m(m+1)/2
for example :-
1+2+3+4+...+18+19+20 = 20(20+1)/2 = 210
my question is :-
what about the series :-
∑ from n = 1 ,until m for n^2 ?
it isn't an artithmetical or even a geometrical series then :-
what kind of series is it ?
and how could they find that :-
∑ from n = 1 ,until m for n^2 = [m(m+1)(2m+1)]/6 ?
for example :-
(1^2)+(2^2)+(3^2)+(4^2)+(5^2)= 5(5+1)(2*5+1)/6 = 55
and thank you