InterviewSolution
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InterviewSolution
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The sum to infinity of $ \frac{1}{7}+\frac{1}{7^{3}}+\frac{2}{7}+\ldots= $1) 162) 7/243) 5/484) 3/16 |
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Answer» 1/7 +2/7^2 +1/7^3 +2/7^4+.........=(1/7 +1/7^3+......)+2(1/7^2 +1/7^4+.....)2 GPs are formedfor both common ratio r=1/7^2since it is infinite series for thissum=a/(1-r)for 1stsum1=(1/7)/(1-1/7^2)=7/48for 2ndsum2=2×(1/7^2)/(1-7^2)=2/48total sum = sum1 + sum2sum=7/48 + 2/48=9/48=0.1875 |
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