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Express-n-1-49-1-n-n-2-1-as-a-b-2-for-some-integers-a-and-b-




Question Number 131484 by bemath last updated on 05/Feb/21
Express Σ_(n=1) ^(49)  (1/( (√(n+(√(n^2 −1)))))) as a+b(√2)  for some integers a and b
Express49n=11n+n21asa+b2forsomeintegersaandb
Answered by benjo_mathlover last updated on 05/Feb/21
Answered by mr W last updated on 05/Feb/21
(1/( (√(n+(√(n^2 −1))))))  =(√(n−(√(n^2 −1))))  =(√(n−(√((n−1)(n+1)))))  =(√(((n−1)/2)+((n+1)/2)−2(√((((n−1)/2))(((n+1)/2))))))  =(√(((√((n−1)/2)))^2 +((√((n+1)/2)))^2 −2(√((((n−1)/2))(((n+1)/2))))))  =(√(((√((n+1)/2))−(√((n−1)/2)))^2 ))  =(√((n+1)/2))−(√((n−1)/2))  Σ_(n=1) ^(49) ((√((n+1)/2))−(√((n−1)/2)))  =Σ_(n=2) ^(50) (√(k/2))−Σ_(k=0) ^(48) (√(k/2))  =(Σ_(k=0) ^(48) (√(k/2))+(√((49)/2))+(√((50)/2))−(√(1/2))−(√(0/2)))−Σ_(k=0) ^(48) (√(k/2))  =(√((49)/2))+(√((50)/2))−(√(1/2))−(√(0/2))  =(7/( (√2)))+5−(1/( (√2)))  =5+(6/( (√2)))  =5+3(√2)
1n+n21=nn21=n(n1)(n+1)=n12+n+122(n12)(n+12)=(n12)2+(n+12)22(n12)(n+12)=(n+12n12)2=n+12n1249n=1(n+12n12)=50n=2k248k=0k2=(48k=0k2+492+5021202)48k=0k2=492+5021202=72+512=5+62=5+32
Answered by Dwaipayan Shikari last updated on 05/Feb/21
(√(n+(√(n^2 −1)))) =(√((n+(√(n^2 −n^2 +1)))/2))+(√((n−(√(n^2 −n^2 +1)))/2))  =(√((n+1)/2))  −(√((n−1)/2))  Σ_(n=1) ^(49) (1/( (√((n+1)/2))−(√((n−1)/2))))=(1/( (√2)))Σ_(n=1) ^(49) (√(n+1))−(√(n−1))  =(1/( (√2)))((√2)−0+(√3)−1+(√4)−(√2)+....+(√(49))−(√(47))+(√(50))−(√(48)))  =(1/( (√2)))(5(√2)+(√(49))−1)=5+3(√2)
n+n21=n+n2n2+12+nn2n2+12=n+12n1249n=11n+12n12=1249n=1n+1n1=12(20+31+42+.+4947+5048)=12(52+491)=5+32