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Let S=Sigma(n=1)^(999) (1)/((sqrt(n)+sqr...

Let `S=Sigma_(n=1)^(999) (1)/((sqrt(n)+sqrt(n+1))(4sqrt(n)+4sqrtn+1))` , then S equals ___________.

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Verified by Experts

The correct Answer is:
9
Given `S=sum_(n=1)^(9999)1/((sqrtn+sqrn(n+1)(root4n+root(4)(n+1)))`
`=sum_(n=1)^(9999)1/((sqrtn+sqrt(n+1))(root4n+root(4)(n+1)))((root4n-root(4)(n+1))/(root4n-root(4)(n+1)))`
`=sum_(n=1)^(9999)((n+1)^(1//4)-n^(1//4))`
`=((2^(1/4)-1)+(3^(1/4)-2^(1/4))+(4^(1/4)-3^(1/4))+....+((9999+1)^(1/4)-(9999)^(1/4)))`
`=(10^(4))^(1/4)-1`
=9
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CENGAGE-PROGRESSION AND SERIES-Exercise (Numerical)
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  10. The value of the sum Sigma(i=1)^(20) i(1/i+1/(i+1)+1/(i+2)+.....+1/(2)...

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  11. The difference between the sum of the first k terms of the series 1^3+...

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  12. The vlaue of the Sigma(n=0)^(oo) (2n+3)/(3^n) is equal to .

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  13. The sum of the infinite Arithmetico -Geometric progression3,4,4,… is .

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  14. Sigma(r=1)^(50)(r^2)/(r^2+(11-r)^2) is equal to .

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  15. If Sigma(r=1)^(50) (2)/(r^2+(11-r^2)), then the value of n is

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  16. Let lt an gt be an arithmetic sequence of 99 terms such that sum of it...

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  17. Find the sum of series upto n terms ((2n+1)/(2n-1))+3((2n+1)/(2n-1))^2...

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  18. Let S=Sigma(n=1)^(999) (1)/((sqrt(n)+sqrt(n+1))(4sqrt(n)+4sqrtn+1)) , ...

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  19. Let S denote sum of the series 3/(2^3)+4/(2^4 .3)+5/(2^6 .3)+6/(2^7 .5...

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  20. The sum (7)/(2^2xx5^2)+13/(5^2xx8^2)+19/(8^2xx11^2)+…10 terms is S, th...

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