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Question Number 143327 by SOMEDAVONG last updated on 13/Jun/21

1/.lim_(n→+∝) ((6+((6+((6+..........+(6)^(1/3) ))^(1/3) ))^(1/3) ))^(1/3) =?

1/.limn→+∝6+6+6+..........+63333=?

Answered by MJS_new last updated on 13/Jun/21

y=((6+((6+...))^(1/3) ))^(1/3) y^3 =6+y y^3 −y+6=0 ⇒ y=2

y=6+6+...33y3=6+yy3−y+6=0⇒y=2

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