Find the sum of the series #1/(2sqrt1 + 1sqrt2) + 1/(3sqrt2 + 2sqrt3) + 1/(4sqrt3 + 3sqrt4) ........ 1/(100sqrt99 + 99sqrt100)#? ( Please solve without differentiation )

2 Answers
George C.
Sep 19, 2017

#9/10#

Explanation:

Note that:

#1/((n+1)sqrt(n)+nsqrt(n+1))#

#= ((n+1)sqrt(n)-nsqrt(n+1))/(((n+1)sqrt(n)+nsqrt(n+1))((n+1)sqrt(n)-nsqrt(n+1)))#

#= ((n+1)sqrt(n)-nsqrt(n+1))/(n(n+1)^2-(n+1)n^2)#

#= ((n+1)sqrt(n)-nsqrt(n+1))/(n(n+1)((n+1)-n))#

#= ((n+1)sqrt(n)-nsqrt(n+1))/(n(n+1))#

#= sqrt(n)/n-sqrt(n+1)/(n+1)#

So:

#1/(2sqrt(1)+1sqrt(2)) + 1/(3sqrt(2)+2sqrt(3)) + 1/(4sqrt(3)+3sqrt(4)) + ... + 1/(100sqrt(99)+99sqrt(100))#

#= (sqrt(1)/1-color(red)(cancel(color(black)(sqrt(2)/2))))+(color(red)(cancel(color(black)(sqrt(2)/2)))-color(brown)(cancel(color(black)(sqrt(3)/3))))+(color(brown)(cancel(color(black)(sqrt(3)/3)))-color(orange)(cancel(color(black)(sqrt(4)/4))))+...+ (color(green)(cancel(color(black)(sqrt(99)/99)))-sqrt(100)/100)#

#= 1/1-10/100#

#= 9/10#

Cesareo R.
Sep 20, 2017

#1-10/100=0.9#

Explanation:

#((k+1)sqrtk-k sqrt(k+1))/(((k+1)sqrtk+k sqrt(k+1))((k+1)sqrtk-k sqrt(k+1)))=1/sqrtk-1/sqrt(k+1)#

then

#sum_(k=1)^99 1/((k+1)sqrtk+k sqrt(k+1)) = sum_(k=1)^99 (1/sqrtk - 1/sqrt(k+1)) = 1/sqrt1-1/sqrt 100 = 1-sqrt100/100=0.9#

Related questions
Impact of this question
2108 views around the world
You can reuse this answer
Creative Commons License