Sum the first two odd numbers, then the first three, the first four, and so on. From these observations could you guess the sum of the first 100 odd numbers?
So the first two odd numbers are
The first two odd numbers are 1 and 3.
1 and 3? It should have stated from zero.
1+3=4
1+3+5=9
1+3+5+7=16...
so the last addition of all odd nnmbers will be 50^2
1) 1+3 = 4
2) 1 + 3 + 5 = 9 = 4+5
3) 1+3+5+7 = 16 = 9+7
4) 1+3+5+7+9 = 25 = 16+9
in general
sum number n+1 = sum n + 2n+1
but notice
1) 4
2) 9
3) 16
4) 25
now that is interesting, let's look at n=5 and n = 6
5) 25+ 11 = 36 LOL
6) 36 + 13 = 49
well we have the sequence of squares of whole numbers
sum n = (n+1)^2
sum 100 = (101)^2 = 10201
ah, here is the Gauss history I was looking for:
When the great Mathematician Gauss (Germany) was about 12 years old, his
teacher asked the whole class to add up all the integers starting from
1 up to 100. All other children started doing long and boring addition
as you did except young Gauss. The teacher asked him "Why you did not do any work ?"
He gave the correct answer directly. Of course, his teacher surprised and
asked his explanation. He said,"Because, 1 plus 100 is 101, 2 plus 99 is 101,
,..,etc. There are 50 such pairs of 101. So, the final sum is 101*50 = 5050."
That is why Gauss is a such great mathematicain shown his talent
when he was a little body.
Certainly, we cannot expect everybody can be as a genuis as Gauss. But, try
to be more lazy to avoid doing any kind of repetitive boring job. I think
when you did the long add-up, you should have noticed that the numbers are all odd and the difference between every two terms is 2. Then stop doing boring addition and try to find some convienient way (or discuss with somebody)to get the better way of solution.
Recall 1 + 3 + 5 + ...+197+199
Use the trick from Gauss. Note 1+199 = 3+197 = 5+195 = ... = 200.
How many such paris , you can see there are 50 pairs[1,3 up to 99].
They are (1,199),(3,197),...(97,103),(99,101). Hence, the sum of the given
series is 200*50 = 10000 as your answer.
Thus, the general form of an Arithmetic Series with leading term
ao and common ratio d as
the nth term a_n = ao + (n-1) d
the partial sum of the first n terms s_n = (ao + a_n)*n/2
[ (first term + nth term)* number of terms / 2]
= (2ao + (n-1)d] * n /2
Goto your given series
1 + 3 + 5 + ...+197+199
ao = 1, d = 2.
a_n= 199 = ao + (n-1) d = 1 + 2(n-1), so n = 100 (total number of terms)
s_100 = (2ao + (n-1)d] * n /2 = (2+ 2(n-1))*n/ 2 = n^2.
(the simplified formula for the sum of first n odd numbers]
Similarly try to get
1 + 2 + 3 + ...+ (n-1) + n = n(n+1)/2
2 + 5 + 8 + ...+ (3n-4) + 3n-1 = n(3n+1)/2
I also wonder what kind of older people around you that nobody you could
ask or discuss when you were solving with long process.
Of course, you should look for some Web sites for more examples about
arithmetic series (or A.P.)
Damon are you a professor. YOur answer was well explain and detail out. Thankyou
By the way I think your teacher phrased this question in a confusing manner. You are usually asked to add the first 100 odd numebers. That starts with ao = 1, a1 = 3, a2 = 5 etc up to afinal = 199
That gives 50 pairs at 200 (or 100 * 200/2) for a sum of 10,000
HOWEVER your question started with the first sum of 1 and 3 or 4 and went up to 4+99*2 = 202
If you used the usual formulation as suggested by Bob Pursley you would get
50*200 = 10,000