let f(1)= 1 and f(n) 2 summation f(r) from r= 1 to n-1.find summation
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1. The classic argument goes like this:
1 + 2 + 3 + · · · + 100
+ 100 + 99 + 98 + · · · + 1
101 + 101 + 101 + · · · + 101
If S is the sum, 2S = 100 · 101 = 10100, so S = 5050.
2. The area of a strip between the circle of radius r and the
circle of radius r + 1 is π(r + 1)2 − πr
2 = (2r + 1)π, which we
can rewrite as (r + 1)π + rπ. Then
x =
1002π − 992π + 982π − 972π + · · · + 22π − 1
2π
10000π
=
100π + 99π + 98π + 97π + · · · + 2π + π
10000π
=
5050π
10000π
= 0.505.
1 + 2 + 3 + · · · + 100
+ 100 + 99 + 98 + · · · + 1
101 + 101 + 101 + · · · + 101
If S is the sum, 2S = 100 · 101 = 10100, so S = 5050.
2. The area of a strip between the circle of radius r and the
circle of radius r + 1 is π(r + 1)2 − πr
2 = (2r + 1)π, which we
can rewrite as (r + 1)π + rπ. Then
x =
1002π − 992π + 982π − 972π + · · · + 22π − 1
2π
10000π
=
100π + 99π + 98π + 97π + · · · + 2π + π
10000π
=
5050π
10000π
= 0.505.
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Answer:
19. c). 3^(m+1)
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