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Let's talk this out and see if we can work it out.
The sum of the first N odd integers means,
1+3+5+7+9+11+...
Where N is how many odd numbers we're adding.
Let's choose numbers for N, and see if we can find a pattern.
N=1 --> 1
(sum of the first odd integer)
N=2 --> 1 + 3 = 4
(sum of the first 2 odd integers)
N=3 --> 1 + 3 + 5 = 9
(sum of the first 3 odd integers)
N=4 --> 1 + 3 + 5 + 7 = 16
Do you notice a pattern yet? Take a look at when N = 2, what is the sum? That's right, 4!
and when N = 3... the sum is 9.
N = 4 the sum is 16....
I see a pattern, do you?
Answer: If you don't, you'll notice that the sum of the first N odd integers is always = N2
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Three odd consecutive integers whose sum is 176?
First of all, the sum of 3 odd numbers always equals an odd number and 176 is even so this is NOT possible.If you did not know this you could let n be the first odd number, n+2 is the next and n+4 is the third.Add them up and we have n+n+2+n+4=176 or3n+6=176 so3n=170Now you see that 170 is NOT divisible by 3 so ONCE again you see that there is NO solution to this problem. That is to say, there are NOT 3 odd consecutive integers whose sum is 176.
How do you find the sum of the first thirty odd integers?
The arithmetic sequence of odd integers is 1, 3, 5, 7, 9, ... where the common difference is 2. The sum of the first thirty odd integers can be found by using the sum formula such as: Sn = n/2[2a1 + (n - 1)d], where a1 = 1, n = 30, and d = 2. So, S30 = (30/2)[(2)(1) + (30 - 1)(2)] = (15)[2 + (29)(2)] = (15)(60) = 900
Why is the sum of 2 consecutive integers always an odd number?
The sum of two even numbers is always an even number. The sum of two odd numbers is always an even number.The sum of an even number and an odd number is always an odd number.Because integers alternate between even and odd, adding two consecutive integers will always result the sum of an odd number and an even number, which, as stated above, will always be an odd number.Another way to look at it:An even number is divisible by 2, therefore, 2*n (where n is an integer) is always an even number.2*n+1 would be the next number after 2*n (so 2*n & 2*n+1 are consecutive integers).If we add these numbers together, we get 4*n+1 = 2(2*n)+1. Since 2(2*n) is an even number, adding 1 makes it an odd number._________________________________________________________________________I'm not going to but in or anything, seriously that was a good answer
The sum of the squares of two consecutive odd integers is 74 find the two integers?
Let the integers be 'n' & 'n+2' Hence Their squares are n^(2) & (n+2)^(2) Their sum is n^(2) + (n+2)^(2) = 74 Expanding n^(2) + n^(2) + 4n + 4 = 74 2n^2 + 4n -70 = 0 n^2 + 2n - 35 = 0 Factor (n - 7)(n + 5) Hence n = '5' & '7' Odd and consecutive.
What is the sum of the first 22 odd numbers?
The sum of a sequence is given by sum = n/2(2a + (n-1)d) where: n = how many a = first number of sequence d = difference between terms of sequence. For the first 22 odd numbers these are: n = 22 a = 1 d = 2 → sum = 22/2(2×1 + (22 - 1)×2)) = 22² = 484 The sum of the first n odd numbers is always n²: sum = n/2(2×1 + (n-1)2) = n/2(1 + (n-1))×2 = n(n) = n²
