SOLUTION: I need to find another way to solve this problem. I would like to be able to use an equation for it.
Find the sum of these consecutive odd numbers:
1 + 3 + 5 + 7 + ... (al
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Question 11513: I need to find another way to solve this problem. I would like to be able to use an equation for it.
Find the sum of these consecutive odd numbers:
1 + 3 + 5 + 7 + ... (all odd numbers from 7-193) 193 + 195 + 197 + 199.
I already know that the answer is 10,000 because I sat here and added up all of the numbers myself, but I need an equation to turn in, and I haven't been able to find one yet. If you could help I would really appreciate it. Thanks- Lisa
Found 2 solutions by askmemath, khwang:
Answer by askmemath(368) (Show Source): You can put this solution on YOUR website!
I am not sure how much you know about progressions. Consecutive odd numbers come under an Arithmetic progression (a.k.a AP).The difference between consecutive terms is same
for e.g 3-1 = 2
5-3 = 2 and so on...
Sum of the first n odd numbers is n^2
Sum of first 2 odd numbers = 1+3 = 4
n^2 = 2^2 = 4
So sum of first 100 numbers is 100^2= 10,000
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
Poor girl:
Patience is a virture but adding up 100 numbers is not only
a boring job but a big torture. Even you used calculator to do it.
What are you going to do if you try to add
100+100+...+100+100 (100 times).
You won't use addition. Right ? Why? Because there is a common property
of such numbers, that is, they are the same. So, you just multiply
100 by 100 to get the sum instead using direct addition.
Hence, before we start to do the addition or any kind of calculation,
we try to look for if there are some common properties among the given
terms.
Here you were given 1 + 3 + 5 + ...+197+199
[No need to show so many terms as you did
1 + 3 + 5 + 7 + ... (all odd numbers from 7-193) 193 + 195 + 197 + 199.
It is long and redundant.]
You should observe that
(i) the series starting from 1 and ending up with 199.
(ii) All the given terms are odd numbers
(iii)The difference of every consecutive two terms is always 2.
So, (ii) and (iii) are the common properties we can take advantage of
them to simplify the given form and get the sum easily.
This kind of series with the same common difference are called
Arithmetic Series defined as: (a_n means a sub n)
We say {a_n } n>=1 is an Arithmetic Series with common difference
d.
if a_n+1 = a_n + d for all n >=1
Now your given series is an Arithmetic Series with
a_1 = 1 and a_n+1 = an + 2 for all n >= 2. (last term = 199)
Similarly 4, 7, 10,..., 91, 94. (d =3)
8,13,18,..,98,103 (d =5)
are also Arithmetic Series.
Now the question is how do we get the sum of an Arithmetic Series?
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.)
Also, try to find something about geometric series.
Try to read and think carefully my remarks above.
Good luck!
Kenny
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