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