CONCEPT:The sum of n terms of an AP with first term a and common difference d is given by: \(\rm {S_n} = \frac{n}{2} \times \left[ {2a + \left( {n - 1} \right)d} \right]\)CALCULATION:Given: 1, 3, 5, 7, .....50th term is an APHere, a = 1 and d = 2Here we have to find the sum of 50 terms of the given AP i.e n = 50? \(\rm {S_{50}} = \frac{50}{2} \times \left[ {2(1) + {(50 - 1)2} } \right]\)? \(\rm {S_{50}} = 25 \times \left[ {2 + 98 } \right]\)? S50 = 2500Hence, the correct option is 2.Concept:The sum of n odd consecutive number = n2Calculation:Here n = 50 The sum of 50 odd consecutive number = (50)2 = 2500