The number 3136 is the sum of the first how many odd whole numbers?

56

how did you solve

so we have

1+3+5+7+... = 3136

let the number of terms of this AS be n
(n/2)(2a +(n-1)d) = Sum(n)
So we have a=1, d=2 , and n = ?
(n/2)(2 + (n-1)2) = 3136
times 2 ....
n(2 + 2n - 2) = 6272
2n^2 = 6272
n^2 = 3136
n = ± 56 , but n has to be a positive whole number,

there are 56 terms

To find the sum of the first how many odd whole numbers that equals 3136, we can use the formula for the sum of an arithmetic series.

The formula to find the sum of an arithmetic series is given by:

Sum = (n/2) * (2a + (n-1)d)

where:
- Sum is the total sum of the series
- n is the number of terms in the series
- a is the first term in the series
- d is the common difference between consecutive terms

In this case, we want to find the number of terms (n) that add up to 3136. Since we are dealing with odd numbers, we know that the common difference (d) between consecutive terms is 2.

Let's substitute the known values into the formula:

3136 = (n/2) * (2 * a + (n-1) * 2)

Simplifying further:

3136 = n * (2a + 2n - 2)

Dividing both sides of the equation by 2:

1568 = n * (a + n - 1)

Now, we need to find two factors of 1568 that differ by 1, which will give us the possible values for n.

When we calculate the prime factorization of 1568, we get: 2 * 2 * 2 * 2 * 7 * 7.

From this, we can see that 1568 has 8 factors.

The possible pairs of factors that differ by 1 are: (1, 1568), (2, 784), (4, 392), (7, 224), (8, 196), (14, 112), (16, 98), and (28, 56).

However, since we are looking for the number of terms in the series, n, we can disregard the solutions where n is less than 1.

Therefore, the first valid pair for n is (7, 224).

Hence, the number 3136 is the sum of the first 7 odd whole numbers.