Viết chương trình python để tính tổng các số nguyên dương của n(n-2)+(n-4)

304_______1_______7 We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ........ (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ........ (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ........ (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ........ (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ........ (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ........ (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/6 = Σ k28We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ........ (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ........ (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ........ (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ........ (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ........ (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ........ (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/6 = Σ k29 300_______0_______1302303304302306

307We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ........ (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ........ (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ........ (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ........ (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ........ (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ........ (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/6 = Σ k23We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ........ (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ........ (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ........ (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ........ (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ........ (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ........ (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/6 = Σ k24 We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ........ (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ........ (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ........ (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ........ (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ........ (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ........ (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/6 = Σ k23304 302303 304

304

304_______0_______7 We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ........ (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ........ (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ........ (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ........ (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ........ (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ........ (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/6 = Σ k23

 

309

# Python3 Program to0____1_______4 # Python3 Program to2

# Python3 Program to3____42_______4

 

# Python3 Program to5

đầu ra

30

Phương pháp 2. O(1)

Chứng minh.

We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ........ (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ........ (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ........ (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ........ (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ........ (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ........ (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2 n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2 n * (n + 1) * (2 * n + 1)/6 = Σ k2

trăn3

# Python3 Program to

# find sum of square

# of first n natural

# numbers

 

# Return the sum of

300

301

302 303

304_______0_______7 # find sum of square7303 # find sum of square7304 302# of first n natural2# numbers1 # of first n natural5# of first n natural2303 301# of first n natural5 303 # Python3 Program to0304 302_______49_______2_______50_______1 3050_______1