Question

find the following sums a) 1+2+3+.......+200, b) 1+3+5+.......199, c)2+4+6+....+200

Answer 1

Step-by-step explanation:

All of the given terms are Arithmatic progressions.

a) 1+2+3+...+200

The first term is 1, common difference= 2-1= 1

using the formula= Last term= first term+ (number of terms-1)Common Difference

200= 1+(n-1)1

199= n-1

n=200.

Therefore using the formula- Sum= (n/2)(2*First term+(n-1)common difference)

sum= 100(2+199(1))

Sum= 100(201)

Sum= 20100

b)1+3+5+...+199

First term is 1, common difference is 3-1= 2

Using the formula stated above,

199= 1+(n-1)2

198= (n-1)2

99= n-1

Therefore n= 100

Using the formula mentioned above,

Sum= 100/2(2+(100-1)2)

Sum= 50(2+198)

Sum= 50(200)

Sum= 10000

c)2+4+6+...+200

First term is 2, common difference is 4-2=2

Using the formula stated above,

200= 2+(n-1)2

198= (n-1)2

n-1= 99

Therefore, n=100

Using the formula mentioned above,

Sum= 100/2(4+(100-1)2)

Sum= 50(4+198)

Sum= 50(202)

Sum= 10100

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Answer 2

Answer:

(199)²

The Id used here is : (a-b)² = a²+b²+2ab

(200-1)²

(200)² + (1)² - 2(200)(1)

40000 + 1 - 400

39601

Therefore (199)² = 39601

Step-by-step explanation: