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Q: Sum of n th odd number?

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There is a surprisingly easy formula for this. Sum of n odd numbers = n2 So the sum of the first 600 odd numbers (starting with 1 as the very first odd number) is 6002 = 360000.

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 two even numbers is always EVENYes.Proof:We can consider an even number as 2*n, and an odd number as 2*n+1.The sum of two even numbers would be 2*n+2*m=2*(n+m), an even number.

Mean = sum of observations/number of observations Median: Order the observations. Of there are an odd number of observations, the median is the middle one. So if there are n observations (where n is odd) then the median is the (n+1)/2 th observation. If n is even, the median is the average of the n/2 th observation and the (n/2 +1) th. Mode: Group the observations. The mode is the value or values that appear the most often. There may be no mode, a single mode or lots of them.

It will always be odd. A proof: Call the even number m, call the odd number p. 'p' is equal to an even number 'n' + 1. Adding m and p is equal to m + n + 1; since m and n are both even, their sum is also even (see the related questions). m + n + 1 is one more than an even number, so it is odd, and therefore an odd number plus an even number is also always odd. QED

The sum of an odd and an even number is odd. Any odd number can be expressed as 2n + 1 (for some integer "n"). Any even number can be expressed as 2m (for another integer, "m"). The sum of the two is 2(m+n) + 1. Since the expression in parentheses is an integer, multiplying it by 2 gives you an even number. Adding 1 makes the entire expression odd.

The formula for the sum of the squares of odd integers from 1 to n is n(n + 1)(n + 2) ÷ 6. EXAMPLE : Sum of odd integer squares from 1 to 15 = 15 x 16 x 17 ÷ 6 = 680

#include<stdio.h>int main () { int sum = 0; for (int odd=3; odd<100; odd+=2) { printf ("%d\n", odd); sum += odd; } printf ("\nSum = %d\n", sum); return 0; }

Sum of two odd numbers is always an even number.e.g. 1 + 7 = 8Explanation:The successor of an odd number is an even number and the successor is obtained by adding 1.Let us assume any odd number, say n.On adding 1(i.e. an odd number) to n we get an even number and again on addition of 1 or addition of 2(an even number) to n we get an odd number. If we continue to add like this we get to know that addition of two odd numbers is always an even number.

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

Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.

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.

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