Question

One more question tonight!
A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely.

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

Answer 1

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

Sk : ?

Answer 2

write the same statemetn as-it-is, just replace n wid k,

Answer 3

i dont know what the hell sk is?

Answer 4

okay but than how can i solve it with k?

Answer 5

we dont need to solve it.
read the question once, it is asking us to JUST WRITE the statement for Sk and Sk+1

Answer 6

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

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

Answer 7

thats it, we're done wid Sk
work Sk+1 now

Answer 8

how do i work sk+1 tho?

Answer 9

put k+1 in place of n

Answer 10

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

Sk+1 : ?

Answer 11

give it a try

Answer 12

so its going to be sk+1(sk+1 + 1) = {sk+1(sk+1 +1)(sk+1 + 2)] / 3

Answer 13

dont put s

Answer 14

its just 1?

Answer 15

and where are 1.2+ 2.3 + 3.4.... in the start?
write it properly/completely

Answer 16

sorry :(

Answer 17

its okay... try if u can :)

Answer 18

See how we wrote the statemetn Sk above,
Sk+1 should look something like that

Answer 19

is it just 1 tho i put in?

Answer 20

k+1?

Answer 21

yes replace n wid k+1

Answer 22

1*2+2*3+3*4+....+ k+1(k+1 + 1) = [k+1(k+1+1)(k+1 +2)] / 3

Answer 23

\(S\color{red}{n}: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + \color{red}{n}(\color{red}{n} + 1) = [\color{red}{n}(\color{red}{n} + 1)(\color{red}{n} + 2)]/3\)

\(S\color{red}{k+1} : 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + (\color{red}{k+1})(\color{red}{k+1} + 1) = [(\color{red}{k+1})(\color{red}{k+1} + 1)(\color{red}{k+1} + 2)]/3\)

Answer 24

Yup !
sumplify

Answer 25

i have to simplify -____-

Answer 26

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

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

Answer 27

flutter my life -___-

Answer 28

f*ck

Answer 29

:|

Answer 30

thats it?

Answer 31

nope

Answer 32

f*ckkkk

Answer 33

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

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

Answer 34

simplify that, then u may deserve some good sleep :)

Answer 35

[(k+4)]

Answer 36

put it in wolfram

Answer 37

i dont know how to use that

Answer 38

this expression :-
(k+1)(k+2)(k+3)

Answer 39

open wolframalpha

Answer 40

i did but is it the alternate forms???

Answer 41

http://www.wolframalpha.com/input/?i=%28k%2B1%29%28k%2B2%29%28k%2B3%29

Answer 42

yea i got the same thing but which one is the answer lol it confuses me

Answer 43

k+6?

Answer 44

Yup ! its in Alternate forms :
2nd line

(k+1)(k+2)(k+3) = k^3+6k^2+11k+6

Answer 45

okay and thats simplified form.. so i put (k+1)(k+2)= k^3+6k^2+11k+6

Answer 46

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

Sk+1 : 1*2+2*3+3*4+....+ (k+1)(k+2) = [(k+1)(k+2)(k+3)] / 3
: 1*2+2*3+3*4+....+ (k+1)(k+2) = [k^3+6k^2+11k+6] / 3

Answer 47

thats right ! ive put the same above :)

Answer 48

u may divide by 3 if u want, if u aren 't sleepy lol

Answer 49

nah thats too much im half asleep typing lol ..

Answer 50

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

Sk+1 : 1*2+2*3+3*4+....+ (k+1)(k+2) = [(k+1)(k+2)(k+3)] / 3
: 1*2+2*3+3*4+....+ (k+1)(k+2) = [k^3+6k^2+11k+6] / 3
: 1*2+2*3+3*4+....+ (k+1)(k+2) = k^3/3+2k^2+11k/3+2

Answer 51

thats completely simplified... have good sleep :)

Answer 52

thank you soo muchh (:

Answer 53

np :D