數學歸納法....唔識呀 , 救命 !!! 好難呀

(a)

For n = 1,
LHS = 1^3 + 2^3 = 1 + 8 = 9 [Focus on the general term (2n)^3, we have to sum until n = 1, i.e. 2^3.]
RHS = (1^2)[(2+1)^2] = (1)(9) = 9
LHS = RHS.

For n = k+1,
1^3 + 2^3 + 3^3 + ... + (2k)^3 + (2k+1)^3 + [2(k+1)]^3
[Be reminded that we have to sum until [2(k+1)]^3. However, between 2k and 2(k+1), there is 2k+1.]

= (k^2)[(2k+1)^2] + (2k+1)^3 + [2(k+1)]^3
[Use the assume for n = k here.]

= [(2k+1)^2][k^2 + (2k+1)] + [2(k+1)]^3
[Consider the "future" RHS [(k+1)^2]{[2(k+1)+1]^2}, we have to factorize and take (k+1)^2 outside.]

= [(2k+1)^2](k^2 + 2k + 1) + [2(k+1)]^3
["Making" (k+1)^2.]

= [(2k+1)^2][(k+1)^2] + (2^3)[(k+1)^3]

= [(k+1)^2][(2k+1)^2 + 8(k+1)]

= [(k+1)^2][(2k+1)^2 + 4(2k+2)]
[Just keep 2k+2 as I am lazy to expand (2k+1)^2. You should know the "lazyness" advantage. ^_^]

= [(k+1)^2][(2k+1)^2 + 4(2k+1+1)]

= [(k+1)^2][(2k+1)^2 + 4(2k+1) + 4]

= [(k+1)^2]{[(2k+1)+2]^2}

= [(k+1)^2]{[(2k+2)+1]^2}

= [(k+1)^2]{[2(k+1)+1]^2}



(b)

We can't use the formula in (a) to solve the sum directly. As you see, the formula in (a) are used to find the sum of the cubes of nature numbers until n. The question in (b) are used to find the sum of the cubes of odd numbers until 95.

Therefore, we can find the sum of the cubes of nature numbers, then minus that of the even numbers.

1^3 + 3^3 + 5^3 + ... + 95^3

= 1^3 + 2^3 + 3^3 + ... + 96^3 - (2^3 + 4^3 + 6^3 + ... + 96^3)

= (48^2){[2(48)+1]^2} - {[(2)(1)]^3 + [(2)(2)]^3 + [(2)(3)]^3 + ... + [(2)(48)]^3}
[As 2n = 96, n = 96/2 = 48.]

= (48^2)(97^2) - [(2^3)(1^3) + (2^3)(2^3) + (2^3)(3^3) + ... + (2^3)(48^3)]

= (48^2)(97^2) - (2^3)(1^3 + 2^3 + 3^3 + ... + 48^3)

= (48^2)(97^2) - (2^3){(24^2)[2(24)+1]^2}
[As 2n = 48, n = 48/2 = 24.]

= (48^2)(97^2) - (2^3)(24^2)(49^2)

= [(48)(97)]^2 - (8){[(24)(49)]^2}
[Lazy??]

= 10614528

Part (a) should be a normal M.I. question. Part (b) is a little bit more tricky but you should learn to identify the pattern, or the relationship between parts (a) and (b).

(a)
To prove
"1³ + 2³ + 3³ + ... + (2n)³ = n² (2n + 1)² "
for any positive integer n.
When n = 1,　L.H.S. = 1³ + 2³ = 9
　　　　　　 R.H.S. = 1 • 3² = 9 = L.H.S.
　i.e. the statement holds when n = 1.
Assume the statement holds when n = k, for some positive integer k,
　i.e. 1³ + 2³ + 3³ + ... + (2k)³ = k² (2k + 1)².
Then, when n = k+1,
　　L.H.S. = 1³ + 2³ + 3³ + ... + (2k)³ + (2k + 1)³ + (2k + 2)³
　　　　　= k² (2k + 1)² + (2k + 1)³ + (2k + 2)³　, by the assumption
　　　　　= (k² + 2k + 1) (2k + 1)² + (2k + 2)³
　　　　　= (k + 1)² (2k + 1)² + 8 (k + 1)³
　　　　　= (k + 1)² [(2k + 1)² + 8 (k + 1)]
　　　　　= (k + 1)² (4k² + 12k + 9)
　　　　　= (k + 1)² (2k + 3)²　= R.H.S.
　i.e. if the statement holds when n = k, it also holds when n = k + 1.
By the principle of mathematical induction,
　1³ + 2³ + 3³ + ... + (2n)³ = n² (2n + 1)²
　for all positive integers n.

(b)
1³ + 3³ + 5³ + ... + 95³
= (1³ + 2³ + 3³ + ... + 96³) - (2³ + 4³ + 6³ + ... + 96³)
= (1³ + 2³ + 3³ + ... + 96³) - 8 (1³ + 2³ + 3³ + ... + 48³)
= 48² (96 + 1)² - 8 • 24² (48 + 1)²
= 21 678 336 - 11 063 808
= 10 614 528.

For n=1
RHS
= n^2(2n + 1)^2
= 1^2 . 3^2
= 9
= 1^3 + 2^3
= LHS

If the proposition is true for n=p, for n=p+1
LHS
= p^2(2p + 1)^2 + (2p+1)^3 + (2p+2)^3
= (p^2 + 2p + 1)(2p + 1)^2+ (2p+2)^3
= (p+1)^2(2p + 1)^2+ 8(p+1)^3
= (p+1)^2 ((2p + 1)^2+ 8(p+1))
= (p+1)^2 (4p^2 + 12p + 9)
= (p+1)^2 (2(p+1)+1)^2
= RHS

So proved.