Question

A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely. Show your work.
Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3

Answer 1

\[\large S_\color{red}{n}:~1\cdot 2+2\cdot 3+3\cdot 4+\cdots+\color{red}{n}(\color{red}{n}+1)=\frac{\color{red}{n}(\color{red}{n}+1)(\color{red}{n}+2)}{3}\]


For \(\large S_\color{red}{k}\) simply replace \(\large \color{red}{n}\) by \(\large \color{red}{k}\) :

\[\large S_\color{red}{k}:~1\cdot 2+2\cdot 3+3\cdot 4+\cdots+\color{red}{k}(\color{red}{k}+1)=\frac{\color{red}{k}(\color{red}{k}+1)(\color{red}{k}+2)}{3}\]

Answer 2

ok so than how do you simplify it

Answer 3

write out \(\large S_{\color{red}{k+1}}\) also

Answer 4

sorry just realised i skipped part how do i write statements sk and sk+1

Answer 5

so sk is what you just wrote replacing sk+1

Answer 6

i wrote \(S_{\color{red}{k}}\) fir you, see if you can write \(S_{\color{red}{k+1}}\)

Answer 7

ok

Answer 8

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

Answer 9

....i hope thats right lol

Answer 10

that almost looks right! but what happened to the "=" symbol ?

Answer 11

ohh i must not have typed it my bad but i know where it goes and im assuming to simplify it i just shrink up all the excess numbers that can be added together

Answer 12

do we have
\[ S_\color{red}{k+1}:~1\cdot 2+2\cdot 3+3\cdot 4+\cdots+(\color{red}{k+1})(\color{red}{k+1}+1)=\frac{(\color{red}{k+1})(\color{red}{k+1}+1)(\color{red}{k+1}+2)}{3}\]

?

Answer 13

yes

Answer 14

let me break it into two pieces
\[ S_\color{red}{k+1}:~1\cdot 2+2\cdot 3+3\cdot 4+\cdots+(\color{red}{k+1})(\color{red}{k+1}+1)\\=\frac{(\color{red}{k+1})(\color{red}{k+1}+1)(\color{red}{k+1}+2)}{3}\]

Answer 15

so we want to simplify the right hand side is it

Answer 16

you dont want to simplify (k+1)(k+1+1)?

Answer 17

\[ \begin{align}S_\color{red}{k+1}&:~1\cdot 2+2\cdot 3+3\cdot 4+\cdots+(\color{red}{k+1})(\color{red}{k+1}+1)\\~\\
&=\frac{(\color{red}{k+1})(\color{red}{k+1}+1)(\color{red}{k+1}+2)}{3}\\~\\
&=\frac{(k+1)(k+2)(k+3)}{3}
\end{align}\]

Answer 18

Oh yes, we can simplify (k+1)(k+1+1) on left hand side as (k+1)(k+2)

Answer 19

ok i thought so

Answer 20

i think your teachers wants you to expand the right hand side

Answer 21

so my answer to this question is
\[S_{k}:1*2+2*3+3*4+...+ k(k+1)=\frac{ [k(k+1)(k+2)] }{ 3 }\]

Answer 22

and
\[S _{k}:1*2+2*3+3*4+...+ k+1(k+2)=\frac{ [k+1(k+2)(k+3)] }{ 3 }\]

Answer 23

there should be a k+1 in the very beggining though

Answer 24

careful, you must put parenthesis around (k+1)

Answer 25

\[S _{k}:1*2+2*3+3*4+...+ \color{red}{(}k+1\color{red}{)}(k+2)=\frac{ [\color{red}{(}k+1\color{red}{)}(k+2)(k+3)] }{ 3 }\]

Answer 26

also i think ur teacher wants you to expand that right hand side

you remember FOIL right ?

Answer 27

(k+1)(k+2)(k+3) = ?

Answer 28

somewhat allthough your gonna have to remind me what it stood for

Answer 29

maybe first work (k+2)(k+3)

Answer 30

use any method that you're familiar wid

Answer 31

just multiply those two binomials

Answer 32

god i used to know this and it irritates me i cant now

Answer 33

haha FOIL is slightly over rated memorization trick

Answer 34

2(k+5)

Answer 35

?

Answer 36

(k+2)(k+3) = k(k+3) + 2(k+3)

Answer 37

now distribute

Answer 38

how did you do that

Answer 39

just distributed

Answer 40

\((k+2)(\color{red}{k+3})\)

think of that red part as a "single" object for now

Answer 41

explain lol i dont remember how you change (k+2) (k+3) to k(k+3) + 2(k+3)

Answer 42

ok

Answer 43

\((k+2)(\color{red}{k+3})\)


first multiply that red part by \(k\), then by \(2\) :
\(k(\color{red}{k+3})+2(\color{red}{k+3})\)

Answer 44

ok than what

Answer 45

then distribute and combine like terms etc

Answer 46

why do you think she wants this

Answer 47

maube just ask her if she really wants this

Answer 48

i would if i could lol she dont answer my emails

Answer 49

would you mind simplifying that further for me so i can see it

Answer 50

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

Answer 51

so 6+11 k+6 k^2+k^3 is what i want

Answer 52

thats the numerator

Answer 53

wolfram's explanation is nice

Answer 54

im confused