Wee Teck's answer to Wee Teck's Junior College 2 H2 Maths Singapore question.
a)
∑ (r +1)2^r / [(r + 2)(r +3)] (last term n, r = 1)
= ∑ (r +1)2^(r + 1 - 1) / [(r + 1 + 1)(r + 1 + 2)]
(last term n, r = 1)
= ∑ ½(r + 1)2^(r + 1) / [(r + 1 + 1)(r + 1 + 2)]
(last term n, r = 1)
= ½ ∑ r2^r / [(r + 1)(r + 2)]
(last term n + 1, r = 2)
= ½ ( (∑ from 1st to (n + 1)th term) - 1st term )
= ½[ 2ⁿ+¹+¹/((n+1) + 2) - 1 - 1(2¹)/((1 + 1)(1 + 2)) ]
= ½ [ 2ⁿ+² /(n + 3) - 1 - 2/6]
= ½ [ 2ⁿ+² /(n + 3) - 4/3 ]
= 2ⁿ+¹ /(n + 3) + (-⅔)
A = 3 , B = -⅔
∑ (r +1)2^r / [(r + 2)(r +3)] (last term n, r = 1)
= ∑ (r +1)2^(r + 1 - 1) / [(r + 1 + 1)(r + 1 + 2)]
(last term n, r = 1)
= ∑ ½(r + 1)2^(r + 1) / [(r + 1 + 1)(r + 1 + 2)]
(last term n, r = 1)
= ½ ∑ r2^r / [(r + 1)(r + 2)]
(last term n + 1, r = 2)
= ½ ( (∑ from 1st to (n + 1)th term) - 1st term )
= ½[ 2ⁿ+¹+¹/((n+1) + 2) - 1 - 1(2¹)/((1 + 1)(1 + 2)) ]
= ½ [ 2ⁿ+² /(n + 3) - 1 - 2/6]
= ½ [ 2ⁿ+² /(n + 3) - 4/3 ]
= 2ⁿ+¹ /(n + 3) + (-⅔)
A = 3 , B = -⅔
b)
∑ ln r² (last term 2n, r = 1)
= ∑ 2 ln r (last term 2n, r = 1)
= 2 ∑ ln r (last term 2n, r = 1)
= 2 ( ln1 + ln2 + ln3 +... + ln(2n - 2) + ln(2n - 1) + ln(2n) )
= 2 ln(1 x 2 x 3 x ... x (2n - 2) x (2n - 1) x 2n)
= 2 ln((2n)!)
∑ ln r² (last term 2n, r = 1)
= ∑ 2 ln r (last term 2n, r = 1)
= 2 ∑ ln r (last term 2n, r = 1)
= 2 ( ln1 + ln2 + ln3 +... + ln(2n - 2) + ln(2n - 1) + ln(2n) )
= 2 ln(1 x 2 x 3 x ... x (2n - 2) x (2n - 1) x 2n)
= 2 ln((2n)!)