lim[1/6+1/62+1/63+...1/6n] is where x tends to infinity is

a)1/5

b)1/6

c)-1/5

d)none of these

limn16+162+163+.....+16n=16+162+163+.....+16n+.......Now this is a geometric progression with , first term = 16 and common ratio=16And sum of infinite GP=a1-r=161-16=15

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it will become an infinite G.P

sum=a/(1-r)

=(1/6)/(1-1/6)

=1/5

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