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

Mathematics
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\[1^3 + 2^3 + 3^3 +.....+ 10^3 = \sum_{1}^{10}N^3\]
$$ \sum \limits_{i=1}^{10} i^3 $$
I never Understood sigma notation Someone please Explain Me =)

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Other answers:

sigma means summation
see what i have written above i.e., answer
$$ \sum $$ 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.
in above expression it is sum of cubes of first 10 natural numbers
fool is right ^^^^^^
What is there on top ,bottom and left ?
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 ;)
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
Could you explain why you write 10 up there ?
10 is the number of terms or number of iterations
\[\sum_{N=1}^{10}N^3 = 1^3 +2^3 + 3^3 + 4^3 +5^3 + 6^3 + 7^3 + 8^3 +9^3 + 10^3\]
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 ?
\[\sum_{n= 1}^{50}(2n-1) = 1+3+5+7+....+99\]
this is arithmetic progression, how you find the number of terms in an A.P ? ;)
Understood =)
is that ok fool bro
using the formula Sn = n/2(a+l), right ?
I would never dare to doubt sheggy ;)
hahahaha buddy i just asked u
But Sheg why is that 50 at the top?
Yes barboat .. plug in the values of a and b and find n
same to you aditi
\[\sum_{1}^{99}\] is'nt it like this
Sorry n =1 at the bottom
No, it's an arithmetic progression .. find the n-th term
a = 1 last term = 99 , d = 2 what is n ?
aditi the total number of first 50 odd natural numbers are there so i had put 50 at the top
how come first 50 ?
aditi you know about arithmetic progression ?
Yup
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
then use it :) and sheggy point of view is also the same .. 1,3,5,7 so what is the 50th odd number ?
i thought the top one was last term and bottom first term
No it is the number of iteration
so, Sn= n/2 (100) ? but what do i write on the Sn side ? i cant solve it otherwise
iteration?
WOW I dont Know Maths =D
number of times you want to execute the operation.
itration amt batao usko
No you know maths .. don't give up so easy :)
bataoo
which one is number of times you want to execute the operation.
hahaha,.............see in simple words it is number of odd numbers that u have to add
aditi in which class u r
Sheg grade 1 =P
number of times you want to execute the operation is upper limit - lower limit..
or Grade 1+1 =P
ok gr8 so u have to work hard
I learnt AP without Sigma
you could write sigma notation in various ways
so, Sn= n/2 (100) ? but what do i write on the Sn side ? i cant solve it otherwise
just purchase this book K.C.Sinha
AP is not sigma .. you will learn sigma probably while doing Riemann sums in definite integral
Ok i understood something =P Thanks Fool and Sheg =D Take My Medals =) here you go
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 :)
thanks but I think you did not understand it ? :/
\[99 = 1 + (n - 1) \times 2\] \[99 -1 = (n - 1) \times 2\] \[98 = (n - 1) \times 2\] \[\frac{98}{2} = (n - 1) \] \[49 = (n - 1)\] \[49+1 = n\] \[50 = n\]
i understood the question barboat asked =)
fool buy books by Dr. K. C. Sinha it will help u alot....the books written by him are simply awesome
books are so heavy , i would rather carry a Laptop =P
@ 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
btw $$1+3+5+7+99 = \sum \limits_{i =-5}^{45} (2n+11)$$ am I right ?
here nth term = 99, a= 1 d = 2 n = ? now plug in these values u will get n
I have read the classics Hall and knight in higher algebra sheggy ;)
isnt the formula Sn=n/2 (2a+(n-1)d) ?
btw indians can also write some classics
sorry buddy I hurt to say but I don't agree . most indian authors plagiarized these classics :(
barboat b4 applying that formula u have to apply nth term formula for finding out number of terms
Do you know abut the famous Kanetkar books for C and datastructure ?
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
I don't know about finance but I haven't found any in my domain ..
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
Aha that's an interesting fact :)
yeah just google out Aswath Damodaran he is real gem.......m dying to meet this finance wizard
Hmm
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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