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For-a-positive-number-n-let-f-n-be-the-value-of-f-n-4n-4n-2-1-2n-1-2n-1-calculate-f-1-f-2-f-3-f-40-

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Question Number 137588 by bramlexs22 last updated on 04/Apr/21
For a positive number n , let f(n) be the value of f(n)=((4n+(√(4n^2 −1)))/( (√(2n+1)) +(√(2n−1)))) calculate f(1)+f(2)+f(3)+...+f(40).
Forapositivenumbern,letf(n)bethevalueoff(n)=4n+4n2−12n+1+2n−1calculatef(1)+f(2)+f(3)+…+f(40).
Answered by bemath last updated on 04/Apr/21
⇔ f(n)=((((√(2n+1)))^2 +((√(2n−1)))^2 +(√((2n+1)(2n−1))))/( (√(2n+1)) +(√(2n−1)))) f(n)= ((((√(2n+1)) )^3 −((√(2n−1)) )^3 )/2) f(1)+f(2)+f(3)+...+f(40) = (((√3^3 )−(√1^3 ))/2) + (((√5^3 )−(√3^3 ))/2) + (((√7^3 )−(√5^3 ))/2) +...+ (((√(81^3 ))−(√(79^3 )))/2) [ telescopy series ] f(1)+f(2)+f(3)+...+f(40) = (((√(81^3 ))−(√1^3 ))/2) = ((9^3 −1)/2) = ((728)/2) = 364
⇔f(n)=(2n+1)2+(2n−1)2+(2n+1)(2n−1)2n+1+2n−1f(n)=(2n+1)3−(2n−1)32f(1)+f(2)+f(3)+…+f(40)=33−132+53−332+73−532+…+813−7932[telescopyseries]f(1)+f(2)+f(3)+…+f(40)=813−132=93−12=7282=364

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