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

- anonymous

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

- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

I never Understood sigma notation
Someone please Explain Me =)

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## More answers

- anonymous

sigma means summation

- anonymous

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

- anonymous

$$ \sum $$
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.

- anonymous

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

- anonymous

fool is right ^^^^^^

- anonymous

What is there on top ,bottom and left ?

- anonymous

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 ;)

- anonymous

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

- anonymous

Could you explain why you write 10 up there ?

- anonymous

10 is the number of terms or number of iterations

- anonymous

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

- anonymous

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 ?

- anonymous

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

- anonymous

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

- anonymous

Understood =)

- anonymous

is that ok fool bro

- anonymous

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

- anonymous

I would never dare to doubt sheggy ;)

- anonymous

hahahaha buddy i just asked u

- anonymous

But Sheg why is that 50 at the top?

- anonymous

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

- anonymous

same to you aditi

- anonymous

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

- anonymous

Sorry n =1 at the bottom

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

how come first 50 ?

- anonymous

aditi you know about arithmetic progression ?

- anonymous

Yup

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

No it is the number of iteration

- anonymous

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

- anonymous

iteration?

- anonymous

WOW I dont Know Maths =D

- anonymous

number of times you want to execute the operation.

- anonymous

itration amt batao usko

- anonymous

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

- anonymous

bataoo

- anonymous

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

- anonymous

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

- anonymous

aditi in which class u r

- anonymous

Sheg grade 1 =P

- anonymous

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

- anonymous

or Grade 1+1 =P

- anonymous

ok gr8 so u have to work hard

- anonymous

I learnt AP without Sigma

- anonymous

you could write sigma notation in various ways

- anonymous

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

- anonymous

just purchase this book K.C.Sinha

- anonymous

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

- anonymous

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

- anonymous

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 :)

- anonymous

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

- anonymous

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

- anonymous

i understood the question barboat asked =)

- anonymous

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

- anonymous

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

- anonymous

@ 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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

btw indians can also write some classics

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Aha that's an interesting fact :)

- anonymous

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

- anonymous

Hmm

- anonymous

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.

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