anonymous
  • anonymous
Write the following series by using the sigma notation. 1+8+27+64+...+1000
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[1^3 + 2^3 + 3^3 +.....+ 10^3 = \sum_{1}^{10}N^3\]
anonymous
  • anonymous
$$ \sum \limits_{i=1}^{10} i^3 $$
anonymous
  • anonymous
I never Understood sigma notation Someone please Explain Me =)

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