Questions
is equal to
n digits n digits
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a
4910n−1
b
49102n−1
c
4910n−12
d
None of these
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Correct option is B
(666…6)=6+6×10+6×102+…+6×10n−1n digits=61+10+102+…+10n−1=6910n−1=2310n−1Similarly, (888…8)=8910n−1 n digitsHence, required sum=4910n−12+8910n−1=49102n−2⋅10n+1+2⋅10n−2=49102n−1Similar Questions
Let an be the nth term of the G.P. of positive numbers. the common ratio is
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