The average is 10. You will find that the average when adding together the first n odd numbers is always equal to n.
Let's call the first odd number x. Since the numbers are consecutive, the second odd number would be x + 2. The sum of the two numbers is 44, so the equation would be x + (x + 2) = 44. Solving this equation yields x = 21, which means the two consecutive odd numbers are 21 and 23.
900
The sum of the first n odd numbers is n squared. 25 x 25 = 625
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.
The answer is one million (the average of the first n odd numbers is always n.)
The number of odd numbers between 38 and n = int[abs(n - 38)/2]
Let the odd numbers be n-4, n- 2, n, n + 2 , n + 4 Hence summing. (n - 4) + (n - 2) + n + ( n + 2) + (n + 4) = 235 5n = 235 (NB The constants add to zero) n = 235/ 5 n = 47 Hence the other four odd numbers are, 43,45,49,51
The set of odd numbers is an arithmetic sequence. Let say that the sequence has n odd numbers where the first term is a1 and the last one is n. The formula to find the sum on nth terms for an arithmetic sequence is: Sn = (n/2)(a1 + an) or Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference that for odd numbers is 2. Sn = (n/2)(2a1 + 2n - 2)
The average is 10. You will find that the average when adding together the first n odd numbers is always equal to n.
Let n = smallest of the odd numbers, then let n+2 = the larger of the two numbers (Remember, 1 is not a prime number.) n+ n+2 = {(2)(7)}2 2n +2 = 142 2n = 196 -2 2n = 194 n = 97 n + 2 = 99
Let's call the first odd number x. Since the numbers are consecutive, the second odd number would be x + 2. The sum of the two numbers is 44, so the equation would be x + (x + 2) = 44. Solving this equation yields x = 21, which means the two consecutive odd numbers are 21 and 23.
There is no such number. Given any odd number, n, the number (n + 2) is a greater odd number. You can go on, for ever, finding larger odd numbers.
The number of odd numbers in the Nth row of Pascal's triangle is equal to 2^n, where n is the number of 1's in the binary form of the N. In this case, 100 in binary is 1100100, so there are 8 odd numbers in the 100th row of Pascal's triangle.
Let 'n' be the first odd number,. Then n+ 2 & n + 4 are the second and third odd number. Hence n + (n+2) + ( n + 4) = 339 Add 3n + 6 = 339 3n = 333 n = 111 n + 2 = 113 n + 4 = 115 Hence the three odd consecutive numbers are 111,113 & 115.
let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34
900