Question

Hey!
I'm thinking of deriving 1+4+9+....+n^2=[n(n+1)(2n+1)]/6 directly no imduction
i was thinking of the series \(X_{n+1}-X_n\)
which give me 3+5+7+9+11+13+......+k
which is the sum of the first n odd number that sums to k^2-1 how will this help. Give me just a hint lol

Answer 1

what last term should be 2k-1
not k lol for the sum of the first odd intergers

Answer 2

that*

Answer 3

that gives you a formula for sum of first n natural numbers

you can use that to find sum of squares of first n natural numbers

Answer 4

\[\large \sum \limits_{k=1}^n (k+1)^2 - k^2 = (n+1)^2 - 1\]

Answer 5

the formula for n natural numbers is n(n+1)/2
the one i was thinking about is sum of n first positive integers

Answer 6

\[\large \sum \limits_{k=1}^n (2k+1) = (n+1)^2 - 1\]

\[\large \sum \limits_{k=1}^n 2k+ \sum \limits_{k=1}^n 1 = (n+1)^2 - 1\]

\[\large \sum \limits_{k=1}^n 2k+ n = (n+1)^2 - 1\]

simplifying a bit you get

\[\large \sum \limits_{k=1}^n k = \dfrac{n(n+1)}{2}\]

Answer 7

use the same trick to find the sum of squares
consider \(\sum\limits_{k=1}^n (k+1)^3 - k^3\) to start with

Answer 8

telescoping gives you \[\sum\limits_{k=1}^n \left((k+1)^3 - k^3\right) = (n+1)^3-1 \]

Answer 9

\[\sum\limits_{k=1}^n \left(k^3+3k^2+3k+1 - k^3\right) = (n+1)^3-1 \]

Answer 10

\[\sum\limits_{k=1}^n \left(3k^2+3k+1\right) = (n+1)^3-1 \]

Answer 11

see if u can take it home

Answer 12

eh so what is started with sum of (Xn+1-Xn) which generated the the sum of first n odd integer is what you used! to find the sum of first n natural number
I thought that this would work for sum of the squares

Answer 13

Exactly!

Answer 14

as you can you can use the same trick to find sum of ANY powers of first n natural numbers

Answer 15

eh i guess from i get it! let me see

Answer 16

the only drawback of this method is that you need to know formulas for ALL the lowest powers

Answer 17

That's it! I got I already have the sum of k and the sum of (1)
I just need to factor the RHD
and push the term of the left to the right leaving sum of k^2

Answer 18

Yep! see if u can find a formula for below if u have time :
\[1^4 + 2^4 + 3^4 + \cdots + n^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\]

Answer 19

this part was not easy to come up with though lol
\(\sum\limits_{k=1}^n \left((k+1)^3 - k^3\right) = (n+1)^3-1\)

Answer 20

actually i think thats the only interesting part here

Answer 21

expand that sum and you will see lot of cancellation..

Answer 22

\[\sum\limits_{k=1}^n \left((k+1)^3 - k^3\right) = (2^3 - 1^3) +(3^3-2^3) + (4^3-3^3) + \cdots + (n+1)^3 - n^3\]

Answer 23

oh yeah there is cancellations
interesting

Answer 24

ok for the other one
i have to start with \(\sum (k+1)^4-k^4=(n+1)^4-1\)

Answer 25

In general
\[\sum\limits_{k=1}^n \left(f(r+1) - f(r)\right) = f(r+1) - f(1)\]

this sum is so interesting that it has a name, its called `telescoping sum`

Answer 26

i just realized that i was approaching the same thing with Xn+1-Xn
taking the sum i just needed to telescope things

Answer 27

yes thats a good start, also u need to find sums of cubes upfront.. :)

Answer 28

oh ok! i get the general idea which what I'm looking thanks alot

Answer 29

\[\sum\limits_{k=1}^n \left((k+1)^3 - (k-1)^3\right) = (n+1)^3+n^3-1\]

\[\sum\limits_{k=1}^n \left((k+1)^3 - (k-1)^3\right) =\sum\limits_{k=1}^n \left(6k^2+2\right)\]

\[6\sum\limits_{k=1}^n k^2+2n= (n+1)^3+n^3-1\]

\[6\sum\limits_{k=1}^n k^2= (n+1)^3+n^3-1-2n=n(n+1)(2n+1)\]

Answer 30

then you don't need to know \[\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}\]

Answer 31

cool^_^ series are really interesting hehe

Answer 32

thats a clever trick to avoid computing \(\sum k\) xD

Answer 33

\(\sum_{i=1}^{ n} (k+1)^4-k^4=(n+1)^4-1\)
this one i wrote above is giving me the sum of k^3

Answer 34

Oh yes because k^4's cancel out on left hand side

Answer 35

right^_^

Answer 36

i suppose for k^4 I should do sum of (k+1)^5-k^5
if the pattern continues hehehe

Answer 37

yes:)

Answer 38

Cool! thanks ^_^

Answer 39

Also notice, using @Zarkon 's trick, you can avoid computing all previous odd powers

Answer 40

\[\sum\limits_{k=1}^n \left((k+1)^5 - (k-1)^5\right) = (n+1)^5+n^5-1\]

\[\sum\limits_{k=1}^n \left(10k^4+20k^2+2\right) = (n+1)^5+n^5-1\]

Answer 41

Eh! i thought that won't work for every case lol

Answer 42

it works always... we can convince it ourselves by playing with sums ..

Answer 43

maybe lets do a quick derivation

Answer 44

hmm ok what derivation

Answer 45

prove \[\sum\limits_{k=1}^n \left((k+1)^5 - (k-1)^5\right) = (n+1)^5+n^5-1 \]

Answer 46

Inductions i guess lol

Answer 47

its a one line proof, trust me

Answer 48

i have to go sleep lol
i will work on this tomorrow!
===========
i was thinking of adding that n^5 to both side and manipulate it in the LHS

Answer 49

taking into count the what we did before of course!

Answer 50

\[\begin{align}\sum\limits_{k=1}^n \left((k+1)^5 - (k-1)^5\right) &= \sum\limits_{k=1}^n \left((k+1)^5 - k^5 + k^5- (k-1)^5\right) \\~\\
&= \sum\limits_{k=1}^n \left((k+1)^5 - k^5\right) + \sum\limits_{k=1}^n \left( k^5- (k-1)^5\right) \\~\\
\end{align}\]

Answer 51

i was trying to do the opposite hehe

Answer 52

\[\begin{align}\sum\limits_{k=1}^n \left((k+1)^5 - (k-1)^5\right) &= \sum\limits_{k=1}^n \left((k+1)^5 - k^5 + k^5- (k-1)^5\right) \\~\\
&= \sum\limits_{k=1}^n \left((k+1)^5 - k^5\right) + \sum\limits_{k=1}^n \left( k^5- (k-1)^5\right) \\~\\
&= \left((n+1)
^5-1\right) +\left( n^5-0\right) \end{align}\]

Answer 53

yeah anything works! have good sleep :)

Answer 54

eh i tried to go the other way, I was almost close
alright thanks for the help
and thanks everyone^_^