Find the sum of the first 50 terms of the sequence, 1,3,5,7,9....?

The given sequence is an arithmetic sequence with a common difference of 2.

The formula to find the sum of the first n terms of an arithmetic sequence is:
Sn = n/2 * (2a + (n-1)d)

where:
Sn = sum of the first n terms
n = number of terms
a = first term
d = common difference

In this case, a = 1, d = 2, and n = 50.

Plugging in the values:
Sn = 50/2 * (2*1 + (50-1)*2)
Sn = 25 * (2 + 49*2)
Sn = 25 * (2 + 98)
Sn = 25 * 100
Sn = 2500

Therefore, the sum of the first 50 terms of the sequence 1,3,5,7,9.... is 2500.

The given sequence is an arithmetic sequence with common difference, d = 2.

The sum of n terms of an arithmetic sequence is given by the formula:

$$S_n = \frac{n}{2} [2a + (n-1)d]$$

where a is the first term and d is the common difference.

Substituting a = 1, d = 2, and n = 50, we get:

$$S_{50} = \frac{(50)}{2} [2(1) + (50-1)2]$$
$$= \frac{(50)}{2} [2 + (49)2]$$
$$= 25 [2 + 98]$$
$$= 25 [100]$$
$$= 2500$$

Therefore, the sum of the first 50 terms of the sequence is 2500.