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barboat

  • 3 years ago

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

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  1. sheg
    • 3 years ago
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    \[1^3 + 2^3 + 3^3 +.....+ 10^3 = \sum_{1}^{10}N^3\]

  2. FoolForMath
    • 3 years ago
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    $$ \sum \limits_{i=1}^{10} i^3 $$

  3. Aditi-Meow
    • 3 years ago
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    I never Understood sigma notation Someone please Explain Me =)

  4. sheg
    • 3 years ago
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    sigma means summation

  5. sheg
    • 3 years ago
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    see what i have written above i.e., answer

  6. FoolForMath
    • 3 years ago
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    $$ \sum $$ 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.

  7. sheg
    • 3 years ago
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    in above expression it is sum of cubes of first 10 natural numbers

  8. sheg
    • 3 years ago
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    fool is right ^^^^^^

  9. Aditi-Meow
    • 3 years ago
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    What is there on top ,bottom and left ?

  10. FoolForMath
    • 3 years ago
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    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 ;)

  11. sheg
    • 3 years ago
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    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

  12. barboat
    • 3 years ago
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    Could you explain why you write 10 up there ?

  13. FoolForMath
    • 3 years ago
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    10 is the number of terms or number of iterations

  14. sheg
    • 3 years ago
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    \[\sum_{N=1}^{10}N^3 = 1^3 +2^3 + 3^3 + 4^3 +5^3 + 6^3 + 7^3 + 8^3 +9^3 + 10^3\]

  15. barboat
    • 3 years ago
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    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 ?

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

  17. FoolForMath
    • 3 years ago
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    this is arithmetic progression, how you find the number of terms in an A.P ? ;)

  18. Aditi-Meow
    • 3 years ago
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    Understood =)

  19. sheg
    • 3 years ago
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    is that ok fool bro

  20. barboat
    • 3 years ago
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    using the formula Sn = n/2(a+l), right ?

  21. FoolForMath
    • 3 years ago
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    I would never dare to doubt sheggy ;)

  22. sheg
    • 3 years ago
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    hahahaha buddy i just asked u

  23. Aditi-Meow
    • 3 years ago
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    But Sheg why is that 50 at the top?

  24. FoolForMath
    • 3 years ago
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    Yes barboat .. plug in the values of a and b and find n

  25. FoolForMath
    • 3 years ago
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    same to you aditi

  26. Aditi-Meow
    • 3 years ago
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    \[\sum_{1}^{99}\] is'nt it like this

  27. Aditi-Meow
    • 3 years ago
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    Sorry n =1 at the bottom

  28. FoolForMath
    • 3 years ago
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    No, it's an arithmetic progression .. find the n-th term

  29. FoolForMath
    • 3 years ago
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    a = 1 last term = 99 , d = 2 what is n ?

  30. sheg
    • 3 years ago
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    aditi the total number of first 50 odd natural numbers are there so i had put 50 at the top

  31. Aditi-Meow
    • 3 years ago
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    how come first 50 ?

  32. FoolForMath
    • 3 years ago
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    aditi you know about arithmetic progression ?

  33. Aditi-Meow
    • 3 years ago
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    Yup

  34. sheg
    • 3 years ago
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    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

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

  36. Aditi-Meow
    • 3 years ago
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    i thought the top one was last term and bottom first term

  37. FoolForMath
    • 3 years ago
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    No it is the number of iteration

  38. barboat
    • 3 years ago
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    so, Sn= n/2 (100) ? but what do i write on the Sn side ? i cant solve it otherwise

  39. Aditi-Meow
    • 3 years ago
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    iteration?

  40. Aditi-Meow
    • 3 years ago
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    WOW I dont Know Maths =D

  41. FoolForMath
    • 3 years ago
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    number of times you want to execute the operation.

  42. sheg
    • 3 years ago
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    itration amt batao usko

  43. FoolForMath
    • 3 years ago
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    No you know maths .. don't give up so easy :)

  44. Aditi-Meow
    • 3 years ago
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    bataoo

  45. Aditi-Meow
    • 3 years ago
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    which one is number of times you want to execute the operation.

  46. sheg
    • 3 years ago
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    hahaha,.............see in simple words it is number of odd numbers that u have to add

  47. sheg
    • 3 years ago
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    aditi in which class u r

  48. Aditi-Meow
    • 3 years ago
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    Sheg grade 1 =P

  49. FoolForMath
    • 3 years ago
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    number of times you want to execute the operation is upper limit - lower limit..

  50. Aditi-Meow
    • 3 years ago
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    or Grade 1+1 =P

  51. sheg
    • 3 years ago
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    ok gr8 so u have to work hard

  52. Aditi-Meow
    • 3 years ago
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    I learnt AP without Sigma

  53. FoolForMath
    • 3 years ago
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    you could write sigma notation in various ways

  54. barboat
    • 3 years ago
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    so, Sn= n/2 (100) ? but what do i write on the Sn side ? i cant solve it otherwise

  55. sheg
    • 3 years ago
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    just purchase this book K.C.Sinha

  56. FoolForMath
    • 3 years ago
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    AP is not sigma .. you will learn sigma probably while doing Riemann sums in definite integral

  57. Aditi-Meow
    • 3 years ago
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    Ok i understood something =P Thanks Fool and Sheg =D Take My Medals =) here you go

  58. FoolForMath
    • 3 years ago
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    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 :)

  59. FoolForMath
    • 3 years ago
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    thanks but I think you did not understand it ? :/

  60. sheg
    • 3 years ago
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    \[99 = 1 + (n - 1) \times 2\] \[99 -1 = (n - 1) \times 2\] \[98 = (n - 1) \times 2\] \[\frac{98}{2} = (n - 1) \] \[49 = (n - 1)\] \[49+1 = n\] \[50 = n\]

  61. Aditi-Meow
    • 3 years ago
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    i understood the question barboat asked =)

  62. sheg
    • 3 years ago
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    fool buy books by Dr. K. C. Sinha it will help u alot....the books written by him are simply awesome

  63. Aditi-Meow
    • 3 years ago
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    books are so heavy , i would rather carry a Laptop =P

  64. sheg
    • 3 years ago
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    @ 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

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

  66. sheg
    • 3 years ago
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    here nth term = 99, a= 1 d = 2 n = ? now plug in these values u will get n

  67. FoolForMath
    • 3 years ago
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    I have read the classics Hall and knight in higher algebra sheggy ;)

  68. barboat
    • 3 years ago
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    isnt the formula Sn=n/2 (2a+(n-1)d) ?

  69. sheg
    • 3 years ago
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    btw indians can also write some classics

  70. FoolForMath
    • 3 years ago
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    sorry buddy I hurt to say but I don't agree . most indian authors plagiarized these classics :(

  71. sheg
    • 3 years ago
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    barboat b4 applying that formula u have to apply nth term formula for finding out number of terms

  72. FoolForMath
    • 3 years ago
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    Do you know abut the famous Kanetkar books for C and datastructure ?

  73. sheg
    • 3 years ago
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    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

  74. FoolForMath
    • 3 years ago
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    I don't know about finance but I haven't found any in my domain ..

  75. sheg
    • 3 years ago
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    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

  76. FoolForMath
    • 3 years ago
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    Aha that's an interesting fact :)

  77. sheg
    • 3 years ago
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    yeah just google out Aswath Damodaran he is real gem.......m dying to meet this finance wizard

  78. FoolForMath
    • 3 years ago
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    Hmm

  79. barboat
    • 3 years ago
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    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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