find the sum of the series 101 + 99 + 97 +
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1
the sum of 101 + 99 + 97 = 297
hope this helps you
hope this helps you
Answered by
0
1 + 2 + 3 + 4 + … + 98 + 99 + 100
Gauss noticed that if he was to split the numbers into two groups (1 to 50 and 51 to 100), he could add them together vertically to get a sum of 101.
1 + 2 + 3 + 4 + 5 + … + 48 + 49 + 50
100 + 99 + 98 + 97 + 96 + … + 53 + 52 + 51
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
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.
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48 + 53 = 101
49 + 52 = 101
50 + 51 = 101
Gauss realized then that his final total would be 50(101) = 5050.
The sequence of numbers (1, 2, 3, … , 100) is arithmetic and when we are looking for the sum of a sequence, we call it a series. Thanks to Gauss, there is a special formula we can use to find the sum of a series:
S is the sum of the series and n is the number of terms in the series, in this case, 100.
Hope this helps!
There are other ways to solve this problem. You can, for example, memorize the formula
This is an arithmetic series, for which the formula is:
S = n[2a+(n-1)d]/2
where a is the first term, d is the difference between terms, and n is the number of terms.
For the sum of the first 100 whole numbers:
a = 1, d = 1, and n = 100
Therefore, sub into the formula:
S = 100[2(1)+(100-1)(1)]/2 = 100[101]/2 = 5050
You can also use special properties of the particular sequence you have.
An advantage of using Gauss' technique is that you don't have to memorize a formula, but what do you do if there are an odd number of terms to add so you can't split them into two groups, for example "what is the sum of the first 21 whole numbers?" Again we write the sequence "forwards and backwards" but using the entire sequence.
1 + 2 + 3 + ... + 19 + 20 + 21
21 + 20 + 19 + ... + 3 + 2 + 1
Now if you add vertically you get
22 + 22 + 22 + ... + 22 + 22 + 22 = 21(22) = 462
But this is twice the sum of the first 21 whole numbers so
1 + 2 + 3 + ... + 19 + 20 + 21 = 462/2 = 231
Natasha, Paul and Penny
Gauss noticed that if he was to split the numbers into two groups (1 to 50 and 51 to 100), he could add them together vertically to get a sum of 101.
1 + 2 + 3 + 4 + 5 + … + 48 + 49 + 50
100 + 99 + 98 + 97 + 96 + … + 53 + 52 + 51
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
.
.
.
48 + 53 = 101
49 + 52 = 101
50 + 51 = 101
Gauss realized then that his final total would be 50(101) = 5050.
The sequence of numbers (1, 2, 3, … , 100) is arithmetic and when we are looking for the sum of a sequence, we call it a series. Thanks to Gauss, there is a special formula we can use to find the sum of a series:
S is the sum of the series and n is the number of terms in the series, in this case, 100.
Hope this helps!
There are other ways to solve this problem. You can, for example, memorize the formula
This is an arithmetic series, for which the formula is:
S = n[2a+(n-1)d]/2
where a is the first term, d is the difference between terms, and n is the number of terms.
For the sum of the first 100 whole numbers:
a = 1, d = 1, and n = 100
Therefore, sub into the formula:
S = 100[2(1)+(100-1)(1)]/2 = 100[101]/2 = 5050
You can also use special properties of the particular sequence you have.
An advantage of using Gauss' technique is that you don't have to memorize a formula, but what do you do if there are an odd number of terms to add so you can't split them into two groups, for example "what is the sum of the first 21 whole numbers?" Again we write the sequence "forwards and backwards" but using the entire sequence.
1 + 2 + 3 + ... + 19 + 20 + 21
21 + 20 + 19 + ... + 3 + 2 + 1
Now if you add vertically you get
22 + 22 + 22 + ... + 22 + 22 + 22 = 21(22) = 462
But this is twice the sum of the first 21 whole numbers so
1 + 2 + 3 + ... + 19 + 20 + 21 = 462/2 = 231
Natasha, Paul and Penny
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