Set M is composed of the positive even integers up to 100. Set N is co

2 cents:

2 + 4 + 6 + … + 100
= 2 (1 + 2 + 3 + … + 50)
= \(2 * \frac{(50*51)}{2} = 50 * 51\)

2 + 4 + 6 + … + 100
This is an arithmetic progression with 50 terms
Hence sum = \(\frac{50}{2} [2*2 + (50 – 1) * 2] = 25 (4 + 49 * 2) = 25 * 2 (2 + 49) = 50 * 51\)

2 + 4 + 6 + … + 100
This expression denotes sum of first 50 even natural numbers
We know, sum of first n even natural numbers is calculated by \(n^2\) + n = n (n + 1)
Hence, sum of this series = 50 (50 + 1) = 50 * 51

1 + 3 + 5 + … + 99
This is an arithmetic progression with 50 terms
Hence, sum of this series = \(\frac{50}{2} [2*1 + (50 – 1) * 2] = 25 (2 + 49 * 2) = 25 * 2 (1 + 49) = 50 * 50 = 50^2\)

1 + 3 + 5 + … + 99
This expression denotes sum of first 50 odd natural numbers
We know, sum of first n odd natural numbers is calculated by \(n^2\)
Hence, sum of this series = \(50^2\)

1 + 3 + 5 + … + 99
= (1 + 2 + 3 + … + 100) – (2 + 4 + 6 + … + 100)
[i.e. sum of first 100 natural numbers – sum of all the even natural numbers till 100]

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