plsz helpp me math Proof: 1 + 2 + 3 + ... + 99 = (99*100)/2?

Question

Proof: 1 + 2 + 3 + ... + 99 = (99*100)/2 please explain also

Answer 1

Proof by induction

1...n is n numbers. Want to show 1+...+n=n(n+1)/2

Base case: n=1

1(1+1)/2=1

Assume 1+...+n=n(n+1)/2 for n, now show it for n+1

1+...+(n+1)
=1+...+n+(n+1)
=n(n+1)/2 +(n+1) (from our assumption)
=(n^2+n)/2+(2n+2)/2
=(n^2+3n+2)/2
=(n+1)(n+2)/2

So we have proved it.

Now plug in n=99:

1+...+99=99*100/2=99*50=5050

Answer 2

this series is an A.P in which sum is given by n(n-1)/2 where n is the no of terms in the sequence

therefore n=100 , Sum= 100(100-1)/2= 100*99/2

Answer 3

The algebra in the inductive step would be simplified if we notice that (a million + 2 + ... + n)^2 = [n(n+a million)/2}^2. Now in case you anticipate that a million^3 + 2^3 + ... + n^3 =[n(n + a million)/2]^2, then a million^3 + 2^3 + ... + n^3 + (n + a million)^3 = [n(n + a million)/2]^2 + (n + a million)^3 = n^2(n + a million)^2/4 + (n + a million)^3 = (n + a million)^2(n^2/4 + n + a million) = (n + a million)^2[(n^2 + 4n + 4)/4] = = [(n + a million)(n + 2)/2]^2.

Answer 4

weeellll.....
if u have a set of numbers, in this case 1-99, 50 is the average.
so 1 + 2 + 3 + ... + 99 = 50+50+50+....+50 (so 99x50)
and (99x100)/2 = 99x50

Problem Solved :D
BEST ANZWER MEEEE XD


P.S. to the other people: if she didn't understand this, dont think she'll understand mathematical terms like n and stuff... dont waste the typing :P

Answer 5

let S=1+2+3-----------------97+98+99

rewrite RHS from last term to first term

S=.....99+98+97..............3+2+1

add LHS and RHS

2S =100+100+100---------100

every term in RHS is 100, so there are 99 100s

so 2S = 99*100

S = (99*100)/2

Answer 6

Sn = t1 + (t1+d) + ( t1 +2d) +... +(tn-d) +tn
Sn = tn + (tn-d) + (tn-2d) +...+(t1+d + t1
2Sn = (t1+tn) + (t1+tn) +(t1+tn) +...+ (t1+tn)
So t1+tn is added n times
2Sn = n(t1+tn)
Sn = n(t1+tn)/2

Sn = Sum of n terms
d = constant difference of two terms
t1 = 1st term
tn = nth term n = number of terms