Question Number 103622 by bobhans last updated on 16/Jul/20

Given a = Σ_(n=1) ^(24) (1/((√(n+1))+(√n))) then the value of a + (1/(log _a (bc)+1)) + (1/(log _b (ac)+1)) + (1/(log _c (ab)+1)) = ?

Answered by OlafThorendsen last updated on 16/Jul/20

(1/((√(n+1))+(√n))) = (√(n+1))−(√n) Then a = Σ_(n=1) ^(24) ((√(n+1))−(√n)) = (√(25)) = 5 5+(1/(log_5 (bc)+1))+(1/(log_b (5c)+1))+(1/(log_c (5b)+1)) 5+(1/(((ln(bc))/(ln5))+1))+(1/(((ln(5c))/(lnb))+1))+(1/(((ln(5b))/(lnc))+1)) 5+((ln5)/(ln(bc)+ln5))+((lnb)/(ln(5c)+lnb))+((lnc)/(ln(5b)+lnc)) 5+((ln5)/(ln(5bc)))+((lnb)/(ln(5bc)))+((lnc)/(ln(5bc))) 5+((ln(5bc))/(ln(5bc))) = 5+1 = 6

Commented bybobhans last updated on 16/Jul/20

cooll

Commented bybemath last updated on 16/Jul/20

wrong sir Σ_(n=1) ^(24) (√(n+1))−(√n) = 4

Commented byOlafThorendsen last updated on 16/Jul/20

exact sir. you are right.

Answered by bemath last updated on 16/Jul/20

a+(1/(log _a (abc))) + (1/(log _b (abc))) +(1/(log _c (abc)))= a+ log _(abc) (abc) = a+1 where a = Σ_(n=1) ^(24) (√(n+1)) −(√n) = 5−1 = 4 then ⇔ a+1 = 4+1 = 5