2007-03-28 00:18:44 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
n(n+1)/2 upto n terms
2007-03-28 00:21:53 · answer #1 · answered by Mein Hoon Na 7 · 2⤊ 0⤋
1 + 2 + 3 +.............+ n = n(n+1)/2
Like if n were 10, we would get
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 10*11/2 = 110/2 = 55
2007-03-28 00:25:00 · answer #2 · answered by curious 4 · 0⤊ 0⤋
the sum of the terms of an Arithematic Progression is given by
[n{2a+(n-1)d}]/2
where
n=number of terms
a=first term
d=common difference=difference of any two consecutive terms
1,2,3,4,5,..... is an Arithematic Progression because sum of any two consecutive terms is constant
so,
a=1
d=2-1=1
let n=n
so 1+2+3+4+5+.........+n=[n{2a+(n-1)d}]/2
=[n{2(1)+(n-1)(1)}]/2
=[n{2+n-1}]/2
=[(n)(n+1)]/2
so the sum !+2+3+4+...... depends on the number of terms you wish
in general the answer to above problem is infinity
2007-03-28 00:27:52 · answer #3 · answered by Daredevil 2 · 0⤊ 0⤋
Use the formula
n(n+1)/2 to get the answer.n stands for number of terms.
2007-03-28 00:31:18 · answer #4 · answered by Ragunandan G 1 · 0⤊ 0⤋
INFINITY
Since we are looking for the sum of an arithmetic series that didnt specify it's limit last term, then it would mean infinity therefore the answer is infinity
2007-03-28 00:25:38 · answer #5 · answered by Lucy 2 · 0⤊ 0⤋
Infinite.
2007-03-28 00:26:12 · answer #6 · answered by Anonymous · 0⤊ 0⤋
1+2+3+.................................. = infinity
:P
2007-03-28 00:24:28 · answer #7 · answered by Sephora C 2 · 0⤊ 0⤋
until wad answer we dun no
2007-03-28 00:45:15 · answer #8 · answered by Professsor Daniel 2 · 0⤊ 0⤋
me
2007-03-28 00:21:12 · answer #9 · answered by DeepBlue 4 · 0⤊ 0⤋
TILL HOW MUCH PLEASE GIVE CORRECT INFORMATION
2007-03-28 00:22:22 · answer #10 · answered by Anonymous · 0⤊ 0⤋