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