Answer to Question #223135 in Real Analysis for Darren Agbor

Question

The sum of the first n terms of a sequence (Un) n € N^* is given by Sn = n(n+1) / n+2. Find

a) The n^th term of the sequence

b) The sum of the terms from the 6^th to the 31^st term inclusive and exclusive.

Answer 1

a)

$S_n=S_{n-1}+a_n, n\geq2$

$a_n=S_n-S_{n-1}=\dfrac{n(n+1)}{n+2}-\dfrac{(n-1)(n-1+1)}{n-1+2}$

$=\dfrac{n(n^2+2n+1-n^2-2n+n+2)}{(n+1)(n+2)}$

$=\dfrac{n(n+3)}{(n+1)(n+2)}, n\geq2$

$a_1=S_1=\dfrac{1(1+1)}{1+2}=\dfrac{2}{3}$

b)

$S_5=\dfrac{5(5+1)}{5+2}=\dfrac{30}{7}$

$S_6=\dfrac{6(6+1)}{6+2}=\dfrac{21}{4}$

$S_{30}=\dfrac{30(30+1)}{30+2}=\dfrac{465}{16}$

$S_{31}=\dfrac{31(31+1)}{31+2}=\dfrac{992}{33}$

The sum of the terms from the 6^th to the 31^st term inclusive

$S_{31}-S_5=\dfrac{992}{33}-\dfrac{30}{7}=\dfrac{5954}{231}$

The sum of the terms from the 6^th to the 31^st term exclusive

$S_{30}-S_6=\dfrac{465}{16}-\dfrac{21}{4}=\dfrac{381}{16}$