Question

Is their a formula for finding finite sum of series with changing differences ?

Answer 1

\[Ex\]
\[1+3+6+10+15+21.....\]

Answer 2

\[3+10+19+30+43...\]

Answer 3

there is two formula for finite series one for arithmetic and one for geometric

Answer 4

i know that but this series is different ,it shows changing difference pattern

Answer 5

yeah i mean there is a pattern but not same difference or same ratio but they have a pattern
is that question you have or you just make that

Answer 6

is their formula for finding the nth term here if its a sequence

Answer 7

if yes then their should be formula for finding summation too

Answer 8

i don't think so
we just have to figure out

Answer 9

for example this one
5,10,17,26,3
pattern for this one 5+5,10+7,17+11 mean we adding odd numbers every time

Answer 10

this is too complex for me

Answer 11

You can recognize that we are adding asequence which known to us 1,2,3,4,......n

Answer 12

yes

Answer 13

@geerky42

Answer 14

observe that \(1+3+6+10+15+21.....=1+(1+2)+((1+2)+3)+(((1+2)+3)+4)+((((1+2)+3)+4)+5).....\)

Answer 15

........ (forgot the continuation lol)

Answer 16

we can see that after taking care of parenthesis
we infinite sum of 1+1+1+1....
2+2+2+2+2....
3+3+3+3+3
and so on

Answer 17

n sums of 1 is just n
n sums of 2 is 2n
n sums of 3 is 3n
and so on
then we would have n+2n+3n+4n+5n+6n........+n^2......
if we factor out n
we would have n(1+2+3+4+5+6.....+n) for first n!
and we already know the sum of \(\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)
so the sum of that series for the first n numbers is
\(n\sum_{i=1}^{n}i=\frac{n^2(n+1)}{2}\)

Answer 18

i guess this is how you would look at it!
but not sure if i'm doing it the right way hehe

Answer 19

let's test this if it is true hehe

Answer 20

does not go right hehe

Answer 21

\(\large\ =\dfrac{n(n+1)(n+2)}{6}\)

Answer 22

where did you come with that
that's for power series \(\sum_{k=1}^{n}k^2\)

Answer 23

this a fun series!
1+1+1+1+....+2+2+2+2+.....+3+3+3+3+.....
what could be the sum of this!
i was confident that i was going in the right path hehehe

Answer 24

\(\large\bf u~~~ can~~~ check~~~ it\)
\(\large\sum _{i=1}^{\infty}\ n^2=\frac{n(n+1)(2n+1)}{6}\)

Answer 25

well it is blowing to infinity, but can we have a finite sum for a finite numbers!

Answer 26

yes

Answer 27

No that's a different case
we have the other series 1+2+3+4...
not \(1^2+2^2+3^2+4^2.....\)

Answer 28

but good to see u tried hard....

Answer 29

tag some other people hehe

Answer 30

no need

Answer 31

at last got it

Answer 32

hmmm, you sure! i think there are people who can help much better
and this stuff are fun to think about

Answer 33

@ganeshie8 , @amistre64

Answer 34

i know they can solve in seconds

Answer 35

yes there is lol

Answer 36

well you question is kinda general!

Answer 37

1+3+6+10+15+21

find the first differences (akin to a first derivative if you will)

1+3+6+10+15+21
2 3 4 5 6

now the second differences are constant, a constant second derivative leads us to a quadratic right?

Answer 38

yes

Answer 39

riemann sums?

Answer 40

we can work this by taking the leading values (1,2,1) and applying them as:

1/0! + 2n/1! + 1[n(n-1)]/2!

and simplify as needed, this is for n>=0

Answer 41

its akin to creating a taylor series

Answer 42

oh i see!

Answer 43

now the second differences are constant, a constant second derivative leads us to a quadratic right?
=========
i didn't understand your line here?

Answer 44

\(\large\ n^2-n+1\)

Answer 45

a discrete formula has rates of change that are called differences

a continuous rate of change are called derivatives

Answer 46

its for nth term

Answer 47

you might have to adjust it if your n starts at 1

Answer 48

if n starts at 1, and we are defined for n starts at 0, then let k=n-1, n=k+1

Answer 49

mathmath333 you should also think about triangular numbers
they did not give you just some random quadratic sequence, the sequence is a nice well known triangular numbers

Answer 50

and how did you decide that is quadratic!

Answer 51

@ganeshie8 i absolutely now this

Answer 52

second sequence looks bit off though

Answer 53

eh! i'm lost now! hehe

Answer 54

but its right

Answer 55

f = x^2
f' = 2x
f'' =2

notice that a quadratic has a constant 2nd derivative

Answer 56

yes! i see that

Answer 57

a constant 2nd differences will be a discrete quadratic

Answer 58

@ganeshie8 if i know the formula for nth term for 2nd series then can i find the summaation of the second series?

Answer 59

we can work a long way:

a0 a1 a2 a3 a4
b0 b1 b2 b3
c c c

bn = b0 + nc



a0 a1 a2 a3 a4
b0+0c b0+1c b0+2c b0+3c



a1 = a0 + b0 + 0c

a2 = a0 + b0 + 0c + b0+1c
= a0 + 2b0 + c(0+1)

a3 = a0 + 3b0 + c(0+1+2)

a4 = a0 + 4b0 + c(0+1+2+3)

an = a0 + nb0 + c(1+2+3+n-1)

Answer 60

forgot a .... inthe c parts lol

Answer 61

yes you can find partial sum for any sequence : \[\large \sum \limits_{n=1}^{k} a_n\]

whats your formula for nth term ?

Answer 62

hmm now it looks way better! thanks @amistre64

Answer 63

if you work enough of them you see a correlation to differential equations

Answer 64

the difference tree also creates a kind of pascal triangle

Answer 65

thats like finding 3 equations and solving the unknowns :
1 3 6 10

let the required nth term formula be "ax^2+bx+c"
you can get 3 equations by using the first 3 terms :

a + b + c = 1
4a + 2b + c = 3
9a + 3b + c = 6

we can solve these, but yes you won't learn anything about the pattern with this method

Answer 66

\(\large\ nth~~term=n^2+4n-2\)

Answer 67

oh right! that looks like a pascal triangle!

Answer 68

i had less practice on differential equations! hehehe

Answer 69

partial sum of first \(k \) terms

\[\large \sum \limits_{n=1}^{k} (n^2 + 4n-2) = \sum \limits_{n=1}^{k}n^2 + 4\sum \limits_{n=1}^{k} n -2\sum \limits_{n=1}^{k} 1 \]

Answer 70

you must be having readymade formulas for sum of squares and sum of first n terms already ?

Answer 71

yes i know few

Answer 72

its time to use them to your advantage

Answer 73

if not, you can always use what we did at the start to develop them :)

Answer 74

\[\large \begin{align} \sum \limits_{n=1}^{k} (n^2 + 4n-2) &= \sum \limits_{n=1}^{k}n^2 + 4\sum \limits_{n=1}^{k} n -2\sum \limits_{n=1}^{k} 1 \\~\\ &= ? + ? + ? \end{align} \]

Answer 75

see if you can simplify and get a nice looking expression for the sum in terms of "k" alone

Answer 76

eh he already mentioned the two sum N^2 and N
the sums of 1's are easy

Answer 77

@mathmath333, just check you replies above^_^

Answer 78

lol i will

Answer 79

here in one place :

\(\large \sum \limits_{n=1}^{k}n^2 = \dfrac{k(k+1)(2k+1)}{6} \)

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

\(\large \sum \limits_{n=1}^{k}1 = k \)

Answer 80

thanks everyone

Answer 81

oh its wrong hehe

Answer 82

whats wrong ?

Answer 83

nothing , i posted a wrong equation

Answer 84

\(\large\bf\ =\dfrac{k(2k^2+15k+1)}{6}\)

Answer 85

at last........sigh

Answer 86

very happy