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waheguru
 3 years ago
Can someone help me solve this, I have no clue where to begin
waheguru
 3 years ago
Can someone help me solve this, I have no clue where to begin

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0write 4 new sequences and see if we can develop patterns or equations for them

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.01 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 :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.01 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the difference tiers of the first column produce an equation as: 1 + 24n + 16n(n1); for n=0,1,2,3,...

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0it looks like the second one can be constructed as: 4(1+8n+4n(n1))

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0we can also prolly expand these out and simplify them before crunching them further

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the 4th column can be considered as: 16(1+3n+n(n1)) and the 3rd as: (9+40n+16n(n2))

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0by adding that all together we get: 1 + 24n + 16n(n1)4(1+8n+4n(n1)) (9+40n+16n(n2))+16(1+3n+n(n1)) which the wolf says simplifies to 16n+4 so the question is: what is the summation of n=0 to 2010 of: 16n+4?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{a}^{b}(16n+4)\] \[\sum_{a}^{b}16n+\sum_{a}^{b}4\] \[16\sum_{a}^{b}n+4\sum_{a}^{b}\] \[16\frac{k(k+1)}{2}+4k\] where k is the number of elements

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0might have my numbers a little mixed about, but i think its good

waheguru
 3 years ago
Best ResponseYou've already chosen the best response.0I dont really understand fully,

waheguru
 3 years ago
Best ResponseYou've already chosen the best response.0how did you get the equation

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.0Here'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.
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