a.maths (mi)

(a) let P(n) be the proposition

forP(1),L.H.S.=1x2=2
R.H.S.=1x2x3/3=2

so P(1) is true

assume that P(k) is true for all +ve intergers
i.e.1*2+3*4+5*6+.....+(2k-1)(2k)={k(k+1)(4k-1)}/3

forP(k+1),1*2+3*4+5*6+.....+(2k-1)(2k)+(2k+1)(2(k+1))
=k(k+1)(4k-1)/3+2(2k+1)(k+1)
=(k+1)(k(4k-1)+6(2k+1))/3
=(k+1)(4k^2+11k+6)/3
=(k+1)(k+2)(4k+3)/3
=(k+1)((k+1)+1)(4(k+1)-1)/3

so if P(k) is true,P(k+1) is also true

so by M.i.P(n) is true for all positive integers n

b(1)1*2+3*4+5*6+...+99*100 =1*2+3*4+5*6+...+(2x50-1)(2x50)
=50(50+1)(4(50)-1)/3
=169150

(2)51*52+53*54+55*56+....+99*100=1*2+3*4+5*6+...+99*100 -(1*2+3*4+5*6+...+49x50)
=169150-25(25+1)(4(25)-1)/3
=147700

為左方便睇，d乘號我用括號寫。

a)
For n=1,
LHS = 2
RHS = [1(1+1)(4-1)]/3 = 2 = LHS
∴ the statement is true for n=1
Assume 1(2)+3(4)+5(6)+...+(2k-1)(2k) = [k(k+1)(4k-1)]/3
1(2)+3(4)+5(6)+...+(2k-1)(2k) + 2(2k+1)(k+1)
= [k(k+1)(4k-1)]/3 + 2(2k+1)(k+1)
= {(k+1)[k(4k-1) + 6(2k+1)]} / 3
= [(k+1)(4k^2 + 11k + 6)] / 3
= [(k+1)(k+2)(4k+3)] / 3
∴ the statement is true for all positive integers n

bi) 1*2+3*4+5*6+...+99*100
= 50(51)(199)/3
= 169150

bii) 51*52+53*54+55*56+....+99*100
= 1*2+3*4+...+99*100 - (1*2+3*4+...+49*50)
= 50(51)(199)/3 - 25(26)(99)/3
= 169150 - 21450
= 147700

Let S(n) be the statement "1*2+3*4+5*6+.....+(2n-1)(2n)={n(n+1)(4n-1)}/3"
for all positive integers n

n=1
LHS=1=RHS
therefore S(n)is ture
assume that S(k) is ture
ie. 1*2+3*4+5*6+.....+(2k-1)(2k)={k(k+1)(4k-1)}/3
we want to prove that S9K+1) also ture
ie. 1*2+3*4+5*6+.....+(2k+1)(2k+2)={(k+1)(k+2)(4k+3)}/3

for n=k+1
LHS= 1*2+3*4+5*6+.....+(2k-1)(2k)+(2k+1)(2k+2)
={k(k+1)(4k-1)}/3+(2k+1)(2k+2).....................by assumption
= 1/3( k(k+1)(4k-1)+3(2k+1)(2k+2))
=1/3(4k^3+15k^2+17k+6)
=1/3(k+1)(4k^2+11k+6)
=1/3(k+1)(k+2)(4k+3)
=RHS

therefore S(k+1)is ture
therefore by MI, all S(n) are ture for all positive integers n