How to find the sum, without adding?
(1)(2+4+6+8+10)
(2)(1+3+5+7+9+11+13)
(3)(2+4+6+8+10+12+14+16+18+20)
The sum cannot be found without some kind of addition somewhere, but ....
look at the last one:
add the first and last ---> 22
add the 2nd and 2nd last ---> 22
add the 3rd and 3rd last ----> 22
how such pairs of 22 can you find?
If you have an odd number of terms, do the same thing, but then don't forget to add the left over middle number,
e.g. for the 2nd
1+13 = 14
3+11 = 14
5+9 = 14
So I have three 14's plus the guy in the middle
sum = 3(14) + 7 = 49
do the first one in the same way.
There is another neat way to add the sum of odd numbers
note the sum of the first two = 1+3 = 4 = 2^2
sum of the first three = 1+3+5 = 9 = 3^2
sum of the first four = ... = 4^2
so...
sum of your first 7 odd numbers = 7^2 = 49 , just as above
Ain't math a lot of fun ??
Sol.(1)There are 5 even numbers
5^2+5=30
Sol.(2)There are 7 odd numbers
7^2=49
Sol.(3)There are 10 odd numbers
10^2+10=100+10=110
To find the sum without actually adding the numbers, you can use a formula for the sum of an arithmetic series, which is given by:
Sum = (n/2)(a + L)
Where:
- Sum represents the sum of the series.
- n is the number of terms in the series.
- a is the first term of the series.
- L is the last term of the series.
Let's apply this formula to each series you've provided:
(1) (1)(2 + 4 + 6 + 8 + 10)
Here, we have the first term (a) as 2 and the last term (L) as 10. The number of terms (n) can be found by dividing the difference between L and a by the common difference, which is 2:
n = (L - a) / d
= (10 - 2) / 2
= 8 / 2
= 4
Therefore, the sum (Sum) can be calculated as:
Sum = (n/2)(a + L)
= (4/2)(2 + 10)
= 2(12)
= 24
So, the sum of the series (1)(2+4+6+8+10) is 24.
(2) (2)(1 + 3 + 5 + 7 + 9 + 11 + 13)
In this series, the first term (a) is 1 and the last term (L) is 13. To find the number of terms (n), we can use the formula as follows:
n = (L - a) / d
= (13 - 1) / 2
= 12 / 2
= 6
Now we can calculate the sum:
Sum = (n/2)(a + L)
= (6/2)(1 + 13)
= 3(14)
= 42
Therefore, the sum of the series (2)(1+3+5+7+9+11+13) is 42.
(3) (3)(2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)
Here, the first term (a) is 2 and the last term (L) is 20. Calculating the number of terms (n):
n = (L - a) / d
= (20 - 2) / 2
= 18 / 2
= 9
Now let's find the sum:
Sum = (n/2)(a + L)
= (9/2)(2 + 20)
= 4.5(22)
= 99
Therefore, the sum of the series (3)(2+4+6+...+20) is 99.
In summary, you can find the sum without adding by using the formula for the sum of an arithmetic series.