A.Maths 2 question

Does the question ask you to prove the statement is correct?
I suppose that you question is "Prove for any positive integer n"
1. Let S(n) be the statement "3 + 3^2 + 3^3 + ... +3^n= 3/2 *(3^n -1)"
When n=1,
L.H.S.=3
R.H.S.=3/2 *(3 -1) =3
So, S(1) is true
Assume that S(k) is true, where k is any positive integer
i.e.3 + 3^2 + 3^3 + ... +3^k= 3/2 *(3^k -1)
when n=k+1
S(k+1)= 3 + 3^2 + 3^3 + ... +3^k +3^(k+1)
=3/2 *(3^k -1) +3^(k+1)
=3/2 * [3^k-1 +2/3 *3^(k+1)]
=3/2 * [3^k -1 +2 * 3^k]
=3/2 * [3 *3^k -1]
=3/2 * [3^(k+1) -1]
So, S(k+1) is true
By MI, S(n) is true for all positive integers n

2.Let S(n) be the statement "1^2+3^2+5^2+...+(2n-1)^2=(n(4(n^2)-1))/3"
When n=1,
L.H.S.=1^2=1
R.H.S.= [1(4 * 1 -1)]/3=1
S(1) is true
Assume that S(k) is true, where k is any positive integer
i.e.1^2+3^2+5^2+...+(2k-1)^2={k[4(k^2)-1]}/3
When n=k+1,
S(k+1)= 1^2+3^2+5^2+...+(2k-1)^2 +(2k +1)^2
={k[4(k^2)-1]}/3 + (2k+1)^2
={k[4(k^2)-1] +3 * (2k+1)^2}/3
={4k^3 -k +3(4k^2+ 4k+1)}/3
={4k^3+12k^2+11k+3}/3
={(k+1)(4k^2+8k+3)}/3
={(k+1)[4k^2+8k+4-1]}/3
={(k+1)[4(k+1)^2-1]}/3
So, S(k+1) is true
By MI, S(n) is true for all positive integers n

設P(n)為命題
3+3^2+3^3+...+3^n=3/2(3^n-1)
當n=1，左方=3
右方=3/2(3^1-1)=3/2(3-1)=3/2(2)=3
所以P(1)成立。
對於任何一個正整數k，假設P(k)成立。
即 3+3^2+3^3+...+3^k=3/2(3^k-1)
則 3+3^2+3^3+...+3^k+3^(k+1)
=3/2(3^k-1)+3^(k+1)
=3/2(3^k-1)+3/2[2/3(3^k*3)]
=3/2[(3^k-1)+2(3^k)]
=3/2(3^k+2(3^k)-1)
=3/2(3(3^k)-1)
=3/2(3^(k+1)-1)
所以P(k+1)成立。
根據數學歸納法，對於所有正整數n，P(n)都成立。
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設P(n)為命題
1^2+3^2+5^2+...+(2n-1)^2=(n(4(n^2)-1))/3
當n=1，左方=1
右方=(1(4(1^2)-1))/3=(4-1)/3=1
所以P(1)成立。
對於任何一個正整數k，假設P(k)成立。
即 1^2+3^2+5^2+...+(2k-1)^2=(k(4(k^2)-1))/3
則 1^2+3^2+5^2+...+(2k-1)^2+(2(k+1)-1)^2
=(k(4(k^2)-1))/3+(2(k+1)-1)^2
=1/3(k(4k^2-1))+1/3(3[(2k+2)-1]^2)
=1/3[4k^3-k+3(2k+1)^2)]
=1/3[4k^3-k+3(4k^2+4k+1)]
=1/3[4k^3+12k^2+11k+3]
=1/3[(k+1)(4(k+1)^2-1)]
所以P(k+1)成立。
根據數學歸納法，對於所有正整數n，P(n)都成立。