F.4 maths-.-.-.-.-.-15marks

(a) Prove that 1 / [(k+1)√k + k√(k+1)] = (1 / √k) - [1 / √(k+1)], where k>0.
L.H.S.
= 1 / [(k+1)√k + k√(k+1)]
= [(k+1)√k - k√(k+1)] / {[(k+1)√k + k√(k+1)] [(k+1)√k - k√(k+1)]}
= [(k+1)√k - k√(k+1)] / {[(k+1)√k]2 – [k√(k+1)]2}
= [(k+1)√k - k√(k+1)] / [(k+1)2k – k2(k+1)]
= [(k+1)√k - k√(k+1)] / {k(k+1)[(k+1) – k]}
= [(k+1)√k - k√(k+1)] / [k(k+1)]
= √k / k - √(k+1) / (k+1)
= 1 / √k – 1 / √(k+1)
= R.H.S.

(b) Hence find the value of the following.
[1 / (2+√2)] + [1 / (3√2+2√3)] + [1 / (4√3+3√4)] + ... + [1 / (100√99+ 99√100)]
= [1 / √1 - 1 / √2] + [1 / √2 - 1 / √3] + [1 / √3 - 1 / √4] + . . . + [1 / √99 - 1 / √100]
= 1 / √1 - 1 / √100
= 1 - 1 / 10
= 9 / 10