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Question-192367




Question Number 192367 by Rupesh123 last updated on 15/May/23
Answered by mehdee42 last updated on 15/May/23
t_r =r−(3/2)⇒t_(r+1) =r−(1/2) & t_(r+2) =r+(1/2)& t_(r+3) =r+(3/2)  ⇒t_r t_(r+1) t_(r+2) t_(r+3) =r^4 −(5/2)r^2 +((25)/(26))  ⇒Σ_(r=1) ^n (√(1+t_r t_(r+1) t_(r+2) t_(r+3) ))=Σ_(r=1) ^n (√(r^4 −(5/2)r^2 +((25)/(16))))  =Σ_(r=1) ^n  (r^2 −(5/4))=((n(n+1)(2n+1))/6)−(5/4)n  ✓
tr=r32tr+1=r12&tr+2=r+12&tr+3=r+32trtr+1tr+2tr+3=r452r2+2526nr=11+trtr+1tr+2tr+3=nr=1r452r2+2516=nr=1(r254)=n(n+1)(2n+1)654n
Commented by Rupesh123 last updated on 15/May/23
Nice, sir!

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