Question

Write the following series by using the sigma notation.
1+8+27+64+...+1000

Answer 1

\[1^3 + 2^3 + 3^3 +.....+ 10^3 = \sum_{1}^{10}N^3\]

Answer 2

$$ \sum \limits_{i=1}^{10} i^3 $$

Answer 3

I never Understood sigma notation
Someone please Explain Me =)

Answer 4

sigma means summation

Answer 5

see what i have written above i.e., answer

Answer 6

$$ \sum $$
This is a capital sigma. It's use is best illustrated by an example:

$$
\sum_{i = 1}^4 \frac{1}{i} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}.
$$

You begin by replacing the index (in this case, $i$) with the first value it takes on (iit's lower bound in this case, 1). You then proceed to the next number and keep doing this replacement until you are at the upper limit (in this case, 4). Finally, you add all these terms up.

Answer 7

in above expression it is sum of cubes of first 10 natural numbers

Answer 8

fool is right ^^^^^^

Answer 9

What is there on top ,bottom and left ?

Answer 10

Fool is always right :P :D btw I have to attribute the answer to Austin Mohr read here http://math.stackexchange.com/questions/81921/weird-e-letter-sigma

I was too tired to type when there is already a very good explanation ;)

Answer 11

bottom it is the range where bottom is the lowest value that N can have and at top u having the highest value that N can have

Answer 12

Could you explain why you write 10 up there ?

Answer 13

10 is the number of terms or number of iterations

Answer 14

\[\sum_{N=1}^{10}N^3 = 1^3 +2^3 + 3^3 + 4^3 +5^3 + 6^3 + 7^3 + 8^3 +9^3 + 10^3\]

Answer 15

for this series, 1+3+5+7+..+99
it would be sigma (2n-1). and at the bottom it is r=1, right ?
how do i get the top number ?

Answer 16

\[\sum_{n= 1}^{50}(2n-1) = 1+3+5+7+....+99\]

Answer 17

this is arithmetic progression, how you find the number of terms in an A.P ? ;)

Answer 18

Understood =)

Answer 19

is that ok fool bro

Answer 20

using the formula Sn = n/2(a+l), right ?

Answer 21

I would never dare to doubt sheggy ;)

Answer 22

hahahaha buddy i just asked u

Answer 23

But Sheg why is that 50 at the top?

Answer 24

Yes barboat .. plug in the values of a and b and find n

Answer 25

same to you aditi

Answer 26

\[\sum_{1}^{99}\] is'nt it like this

Answer 27

Sorry n =1 at the bottom

Answer 28

No, it's an arithmetic progression .. find the n-th term

Answer 29

a = 1 last term = 99 , d = 2 what is n ?

Answer 30

aditi the total number of first 50 odd natural numbers are there so i had put 50 at the top

Answer 31

how come first 50 ?

Answer 32

aditi you know about arithmetic progression ?

Answer 33

Yup

Answer 34

and the thing which fool is saying that is also another method which is mostly used............and the thing which i m saying as we are having small size so we can calculate easily

Answer 35

then use it :) and sheggy point of view is also the same .. 1,3,5,7 so what is the 50th odd number ?

Answer 36

i thought the top one was last term and bottom first term

Answer 37

No it is the number of iteration

Answer 38

so, Sn= n/2 (100) ? but what do i write on the Sn side ? i cant solve it otherwise

Answer 39

iteration?

Answer 40

WOW I dont Know Maths =D

Answer 41

number of times you want to execute the operation.

Answer 42

itration amt batao usko

Answer 43

No you know maths .. don't give up so easy :)

Answer 44

bataoo

Answer 45

which one is number of times you want to execute the operation.

Answer 46

hahaha,.............see in simple words it is number of odd numbers that u have to add

Answer 47

aditi in which class u r

Answer 48

Sheg grade 1 =P

Answer 49

number of times you want to execute the operation is upper limit - lower limit..

Answer 50

or Grade 1+1 =P

Answer 51

ok gr8 so u have to work hard

Answer 52

I learnt AP without Sigma

Answer 53

you could write sigma notation in various ways

Answer 54

so, Sn= n/2 (100) ? but what do i write on the Sn side ? i cant solve it otherwise

Answer 55

just purchase this book K.C.Sinha

Answer 56

AP is not sigma .. you will learn sigma probably while doing Riemann sums in definite integral

Answer 57

Ok i understood something =P Thanks Fool and Sheg =D
Take My Medals =) here you go

Answer 58

but I knew back from standard VII or VIII during olympaid training and all, however I was never good then :P also no need to buy any book .. just follow OCW it's great resource :)

Answer 59

thanks but I think you did not understand it ? :/

Answer 60

\[99 = 1 + (n - 1) \times 2\]
\[99 -1 = (n - 1) \times 2\]
\[98 = (n - 1) \times 2\]
\[\frac{98}{2} = (n - 1) \]
\[49 = (n - 1)\]
\[49+1 = n\]
\[50 = n\]

Answer 61

i understood the question barboat asked =)

Answer 62

fool buy books by Dr. K. C. Sinha it will help u alot....the books written by him are simply awesome

Answer 63

books are so heavy , i would rather carry a Laptop =P

Answer 64

@ Barboat what is the formula for nth term of AP
\[t_{n} = a + (n -1)d\]

where \[t_{n}\] is the nth term
a first term
d common difference
n number of terms

Answer 65

btw $$1+3+5+7+99 = \sum \limits_{i =-5}^{45} (2n+11)$$ am I right ?

Answer 66

here nth term = 99,
a= 1
d = 2
n = ?
now plug in these values u will get n

Answer 67

I have read the classics Hall and knight in higher algebra sheggy ;)

Answer 68

isnt the formula Sn=n/2 (2a+(n-1)d) ?

Answer 69

btw indians can also write some classics

Answer 70

sorry buddy I hurt to say but I don't agree . most indian authors plagiarized these classics :(

Answer 71

barboat b4 applying that formula u have to apply nth term formula for finding out number of terms

Answer 72

Do you know abut the famous Kanetkar books for C and datastructure ?

Answer 73

i too carry the same feelings as u but not in case of K. C. Sinha.
and in case of finance books i never refer indian authors. I prefer to read other than indian publication house books

Answer 74

I don't know about finance but I haven't found any in my domain ..

Answer 75

yeah i had found in fiance domain but the master of finance field is also from india and whole world is reading his books only and due to whom i had been to this website

Answer 76

Aha that's an interesting fact :)

Answer 77

yeah just google out Aswath Damodaran he is real gem.......m dying to meet this finance wizard

Answer 78

Hmm

Answer 79

mm for the 1+8+27+64+...+1000 series, we can use the Tn=a+(n-1)d formula to find the top number ? but the common difference isnt the same.