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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
What is the formula to find sum of n odd numbers?
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)
What does d2 mean in algebra?
That would mean d times 2.
What is the sum of the first 45 consecutive odd numbers?
You can use the formula for the sum of an arithmetic series. Sn = (n/2)[2a + (n - 1)d].....n is the number of terms, a is the first term and d the common difference between terms. In this question, n = 45, a = 1 (as 1 is the first odd number) and d = 2 (the difference between consecutive odd numbers). Then, S(45) = (45/2)[2 + 44x2] = (45/2) x 90 = 2025.
What is the sum of the first 90 consecutive odd numbers?
Use the equation S/2 = a + ( n - 1)d Were a = 1 d = 2 (Consecutive odd numbers havie a difference of '2' n = 90 Substitute S = 2[(1) + (90 - 1) 2[ S = 2 + 2(89)2 S = 2 + 356 S = 358
