Update 1: When n = 1

L.H.S. = 1？ + 2? = 9 R.H.S. = 1？ 3? Can you explain it to me? ...

Update 2: L.H.S. = 1? + 2? + 3? +...+ (2k)？ + (2k + 1)？ + (2k + 2)？ E is not clear at all.

The horse can explain it in detail

(a) for n = 1

LHS =13+2 3 =1+8 = 9 [Focus on the general term (2n)^3

We must sum until n = 1

That is 2 3.] RHS = (12) [(2+1) 2] = (1) (9) = 9lhs = RHS. For n = k+ 1

13+2 3+3+...+(2k) 3+(2k+1) 3+[2 (k+1)] 3 [reminds us that we must sum until [2 (k+)

Beeen 2k and 2(k+ 1)

There are 2k+ 1. ] = (k 2) [(2k+1) 2)+(2k+1) 3+[2 (k+1)] 3 [The assumption of n = k is used here. ] = [(2k+1) 2] [k 2+(2k+1)]+[2 (k+1)] 3 [Think about the future & ampquot [(k+/kloc-)

We must factorize and exclude k+1) 2. ]=[(2k+ 1)^2](k^2+2k+ 1)+[2(k+ 1)]^3[& amp； Quot Making & quot (k+1) 2.] = [(2k+1) 2] [(k+1) 2]+(2 3) [(k+1)] Kloc-0/) 2] [(2k+1) 2+4 (2k+2)] [Just keep 2k+2 for my lazy expansion (2k+1) 2. You should know & ampquot laziness and laziness. Quot 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+ / Kloc-0/)+2] 2} = [(k+1) 2] {[(2k+2)+1] 2} = [(k+1) 2] {[2 (k) Do not use the formula in (a) to calculate the total directly. As you can see.

The formula in (a) is used to find the sum of cubes of natural number n, and the problem in (b) is used to find the sum of cubes of odd number 95. therefore

We can find the sum of the cubes of natural numbers.

Then subtract the even number. 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)( / kloc-0/)]^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] =

Part (a) should be a normal M.I. problem. Part (b) is a bit complicated, but you should learn to recognize patterns.

Or the relationship between part (a) and part (b). Proof & quot1? + 2? + 3? +...+ (2n)？ = n？ (2n + 1)？ & ampquot For any positive integer n. When n = 1

L.H.S. = 1？ + 2? = 9 R.H.S. = 1？ 3? = 9 = L.H.S.i That is, when n = 1 Assume that the statement holds when n = k.

For a positive integer k

That is 1? + 2? + 3? +...+ (2k)？ = k？ (2k + 1)？ . then

When n = k+ 1

L.H.S. = 1？ + 2? + 3? +...+ (2k)？ + (2k + 1)？ + (2k + 2)？ = k？ (2k + 1)？ + (2k + 1)？ + (2k + 2)？

By assuming = (k? + 2k + 1) (2k + 1)？ + (2k + 2)？ = (k + 1)？ (2k + 1)？ + 8 (k + 1)？ = (k + 1)？ [(2k + 1)？ + 8 (k + 1)] = (k + 1)？ (4k？ + 12k + 9) = (k + 1)？ (2k + 3)？ = R.H.S.i That is, when n = k, does this statement hold?

It also holds when n = k+1 According to the principle of mathematical induction

1? + 2? + 3? +...+ (2n)？ = n？ (2n + 1)？ For all positive integers n. (b) 1? + 3? + 5? +...+ 95? = ( 1? + 2? + 3? +...+ 96? ) - (2? + 4? + 6? +...+ 96? ) = ( 1? + 2? + 3? +...+ 96? ) - 8 ( 1? + 2? + 3? +...+ 48? ) = 48? (96 + 1)? - 8 ? 24? (48 + 1)? = 2 1 678 336 - 1 1 063 808 = 10 6 14 528.

For n =1RHS = n 2 (2n+1) 2 =12. 3 2 = 9 =13+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(p