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waheguru

Can someone help me solve this, I have no clue where to begin

  • one year ago
  • one year ago

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  1. waheguru
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    • one year ago
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  2. amistre64
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    write 4 new sequences and see if we can develop patterns or equations for them

    • one year ago
  3. amistre64
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    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 4 9 16 25 36 49 64 81 100 121 144 might be easier to construct in excel :)

    • one year ago
  4. amistre64
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    1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961 1024 1089 1156 1225 1296 1369 1444 1521 1600 1681 1764 1849 1936 2025 2116 2209 2304 2401 2500 2601 2704 2809 2916 3025 3136 3249 3364 3481 3600 3721 3844 3969 4096 4225 4356 4489 4624 im sure theres a way to develop an explicit equation for each column then run sum summation rules on them

    • one year ago
  5. amistre64
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    the difference tiers of the first column produce an equation as: 1 + 24n + 16n(n-1); for n=0,1,2,3,...

    • one year ago
  6. amistre64
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    it looks like the second one can be constructed as: -4(1+8n+4n(n-1))

    • one year ago
  7. amistre64
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    we can also prolly expand these out and simplify them before crunching them further

    • one year ago
  8. amistre64
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    the 4th column can be considered as: 16(1+3n+n(n-1)) and the 3rd as: -(9+40n+16n(n-2))

    • one year ago
  9. amistre64
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    by adding that all together we get: 1 + 24n + 16n(n-1)-4(1+8n+4n(n-1)) -(9+40n+16n(n-2))+16(1+3n+n(n-1)) which the wolf says simplifies to 16n+4 so the question is: what is the summation of n=0 to 2010 of: 16n+4?

    • one year ago
  10. amistre64
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    \[\sum_{a}^{b}(16n+4)\] \[\sum_{a}^{b}16n+\sum_{a}^{b}4\] \[16\sum_{a}^{b}n+4\sum_{a}^{b}\] \[16\frac{k(k+1)}{2}+4k\] where k is the number of elements

    • one year ago
  11. amistre64
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    might have my numbers a little mixed about, but i think its good

    • one year ago
  12. waheguru
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    I dont really understand fully,

    • one year ago
  13. waheguru
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    how did you get the equation

    • one year ago
  14. JamesJ
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    Here's a different approach. Observe that the nth and the (n+1)th square differ by 2n + 1, as (n+1)^2 - n^2 = 2n + 1. For example 4 and 9 difference by 2x2 + 1 = 5 9 and 16 by 2x3 + 1 = 7 etc. Now, looking at the sequence you have, let's look at each group of four and write down an equation for it, starting with -4 -9 16 25 -4 and 16 differ by and add up to (2n + 1) + (2n + 3) = 4n + 4,where n = 2 -9 and 25 differ by and add up to (2(n+1) + 1) + (2(n+1) + 3) = 4n + 8 (n = 2) Hence -4 - 9 + 16 + 25 = 8n + 12 = 28 (n = 2) Now take this term 8n + 12 and generalize as needed for your problem.

    • one year ago
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