A.Maths Question,Mathematical Induction!(20 Marks)!!要今日完成!!!

2008-10-11 10:21 pm

Prove,by mathematical induction,that the following propositions are true for all positiive integers n.

1)3+3²+3³+...+3^n=3/2(3^n-1)

2)1‧4+2‧7+3‧10‧+...+n(3n+1)=n(n+1)²

3)1²+3²+5²+...+(2n-1)²=n(4n²-1)/3

4)1‧2‧3‧+2‧3‧4‧+3‧4‧5‧+...+n(n+1)(n-2)
=1/4n(n+1)(n+2)(n+3)

5)3/4 + 5/36 + 7/144 + 2n+1/n²(n+1)² = 1 - 1/(n+1)²

6)1/1 x 2 + 1/2 x 3 + ...+ 1/n(n+1) = n/n+1

7)1/1 x 4 + 1/4 x 7 + 1/7 x 10 +...+ 1/(3n-2)(3n+1)
=n/3n+1

Prove,by mathematical induction,that the following propositions are true for all natural numbers n.

8)1/ 2‧5‧8 + 1/ 5‧8‧11 + 1/ 8‧11‧14 +...+ 1/ (3n-1)(3v+2)(3n+5)=1/60 - 1/ 6(3n+2)(3n+5)

更新1:

如果嫌長的話...可以由"When n=k+1"開始寫solution,,,thanks!!!

回答 (1)

碧咸仔

2008-10-12 4:09 am

✔ 最佳答案

1)3+3²+3³+...+3^n=3/2(3^n-1)

n=k+1

L.H.S=
=(3/2) * (3^k-1) + 3^(k+1)
=(3/2) * (3^k-1) + 2*3^(k+1)/2
=(1/2)*[(3 * (3^k-1) + 2*3^(k+1)]
=(1/2)*[(3 * (3^k-1) + 2*3*3^k]
=(1/2)*3*{ (3^k-1)+ 2*3^k}
=3/2[3^(k+1)-1]

ps : 3^k + 2*3^k = 3*3^k = 3^(k+1) [by laws of indices]

R.H.S=
3/2[3^(k+1)-1]

2)1‧4+2‧7+3‧10‧+...+n(3n+1)=n(n+1)²

1*4+2*7+3*10+....+n(3n+1)=n(n+1)^2

when n = k+1
R.H.S
=(k+1)(k+2)^2
L.H.S
= k(k+1)^2+(k+1)(3k+4)
= (k+1)[k*k+1+3k+4]
= (k+1)(k^2+k+3k+4)
= (k+1)(k^2+4k+4)
= (k+1)(k+2)^2

2008-10-11 20:24:25 補充：
3)1²+3²+5²+...+(2n-1)²=n(4n²-1)/3

when n=k+1

R.H.S = (k+1)[4*(k+1)^2-1]/3 = (k+1) * [ 4k^2+8k + 4-1]/3

= (k+1)*(4k^2+8k+3)/3

=(4k^3+12k^2+3+11k)/3

L.H.S

= k*(4k^2-1)/3 + [2(k+1)-1]^2

= k*(4k^2-1)/3 + [2k+2-1]^2

= [ k*(4k^2-1) + 3(2k+1)^2 ]* [1/3]

= [4k^3-k+3(4k^2+1+4k)]*(1/3)

=(4k^3+12k^2+11k+3)/3

👍 0 👎 0