F.4 Maths Module 2 Question

Q1,1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^2

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Assime that P(k) is ture, i.e. 1^3+2^3+3^3+...+k^3 = (1+2+3+...+k)^2
for n = k+1
1^3+2^3+3^3+...+k^3 +(k+1)^3
= (1+2+3+...+k)^2 +(k+1)^3
= [k(k+1)/2]^2 + (k+1)^3
= (k+1)^2 [k^2/4 + (k+1)]
= (k+1)^2[k^2 +4k +4 ]/4
= (k+1)^2(k+2)^2/2^2
= [(k+1)(k+2)/2]^2
= [1+2+3+...+k+(k+1)]^2
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..Q2,
1-1/2+1/3--1/4+...+1/2n-1 -1/2n = 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n
let P(n) be the statment.
1-1/2+1/3--1/4+...+1/2n-1 -1/2n = 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n
for n = 1
L.H.S. = 1-1/2 = 1/2
R.H.S. = 1/2(1) = 1/2
therefore P(1) is true.
Assume that P(k) is true
1-1/2+1/3--1/4+...+1/2k-1 -1/2k = 1/k+1 + 1/k+2 + 1/k+3 + ... + 1/2k
for n = k+1
1-1/2+1/3--1/4+...+1/2k-1 -1/2k + 1/2k+1 - 1/2k+2
= 1/k+1 + 1/k+2 + 1/k+3 + ... + 1/2k + 1/2k+1 - 1/2k+2
= 1/k+2 + 1/k+3 + ... + 1/2k + [2/2k+2 + 1/2k+1 - 1/2k+2]
= 1/k+2 + 1/k+3 + ... + 1/2k + + 1/2k+1 + 1/2k+2
hence P(k+1) is ture if P(k) is true

by the principle .....
Q3(a)
a1 = (1+1/2)-1 = 1/2
a2 = (1+1/2)(1+1/3)-1 = 2/2
a3 = (1+1/2)(1+1/3)(1+1/4)-1 = 3/2
a4= (1+1/2)(1+1/3)(1+1/4)(1+1/5)-1 = 4/2
(b)
an = n/2
(c)let P(k) be the statement
an = n/2, ak= (1+1/2)(1+1/3)(2+1/4)...(1+1/k+1)-1
.by answer (b)
a1 = (1+1/2)-1 = 1/2
a2 = (1+1/2)(1+1/3)-1 = 2/2
Assume that P(k) is true
ak = k/2
a(k+1) = (1+1/2)(1+1/3)(2+1/4)...(1+1/k+1)(1 + 1/k+2)-1
= (k/2 +1)(1 + 1/k+2) - 1
= [(k+2)(k+3)/2(k+2) ] - 1
= (k^2 +3k +2)/2(k+2)
= (k+1)/2
hence, P(k+1) is true if P(k) is true, by the principle of the mathematical induction ...
註:Q1,1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^2
not ,1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^3

Q2,1-1/2+1/3--1/4+...+1/2n-1 -1/2n = 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n
1-1/2+1/3--1/4 can you use (..) to show it clear- -?