Question

A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true.

Sn: 1^2 + 4^2 + 7^2 + . . . + (3n - 2)^2 = n(6n^2-3n-1)/2

Answer 1

here is an example:

http://openstudy.com/study#/updates/5179f862e4b0c3f42e512b72

Answer 2

@satellite73 Help me Please

Answer 3

@Hero @TuringTest

Answer 4

ok wat r u asking for

Answer 5

A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Sn: 1^2 + 4^2 + 7^2 + . . . + (3n - 2)^2 = n(6n^2-3n-1)/2

Answer 6

well good news i can help

Answer 7

for S1 just plug n = 1 into the formula and see if it works out to 1^2 ( 1)

Answer 8

S1 = 1(6*1^2 - 3(1) - 1) / 2

Answer 9

ok that 1 * (6 - 3 -1) / 2 work out the brackets first = 1 * 2 / 2 = 2 / 2 = 1 which checks out

Answer 10

ok so do the same thing for S2 ( n = 2) and the result should equal the sum of 2 terms - that -s 1^2 + 4^2 which is 1 + 16 = 17

Answer 11

that is 1^1 ^ 4^2 + 7^2 = 1 + 16 + 49 which equlas 66

Answer 12

do u understand it now @Yacoub1993

Answer 13

so we just replace n with the digits

Answer 14

yes

Answer 15

what are the ..... for

Answer 16

7^2 + . . . + (3n - 2)^2 the dots that are inbetween

Answer 17

yes

Answer 18

yes what.... what are the dots in between for

Answer 19

u can take it out if u want

Answer 20

A statement Sn about the positive integers is given. Write statements \[S _{k}\] and \[S_{k+1}\], simplifying\[ S _{k=1}\] completely.

Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3

Answer 21

what about this one @TheRealMeeeee

Answer 22

let me see

Answer 23

Sk: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + k(k + 1) = [k(k + 1)(k + 2)]/3

Answer 24

Sk+1: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + (k+1)(k+1 + 1) = [(k+1)(k+1 + 1)(k+1 + 2)]/3

Answer 25

Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + (k+1)(k+1 + 1) = [(k+1)(k+1 + 1)(k+1 + 2)]/3 Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + (k+1)((k+1) + 1) = [(k+1)(k+1 + 1)(k+1 + 2)]/3 Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + k((k+1) + 1)+1((k+1) + 1) = [(k+1)(k+1 + 1)(k+1 + 2)]/3 Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + k(k+1) + k + 1(k+1) + 1 = [(k+1)(k+1 + 1)(k+1 + 2)]/3

Answer 26

So this proves that Sn: 1 · 2 + 2 · 3 + 3 · 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3 for all natural numbers n.