One way to work out the problem above is to make use of arithmetic sequences.
The bottom row is growing by two (arithmetic sequence) by the time it gets to case 100 the bottom row will have 3 + 2(100-1) objects, which is 201.
This means that in case 100 the rows should follow the following pattern beginning with the first row all the way up to the last row: (1 +3 +...+201)
To figure out the summation one can use the formula (n(a1+an))/2
But first,
To figure out n: 201=1 +2(n-1) n=101 So the sum would be (101(1+201))/2 10,201
For detailed explanations on arithmetic sequences and the formulas check out this website:
http://www.regentsprep.org/Regents/math/algtrig/ATP2/ArithSeq.htm
The bottom row is growing by two (arithmetic sequence) by the time it gets to case 100 the bottom row will have 3 + 2(100-1) objects, which is 201.
This means that in case 100 the rows should follow the following pattern beginning with the first row all the way up to the last row: (1 +3 +...+201)
To figure out the summation one can use the formula (n(a1+an))/2
But first,
To figure out n: 201=1 +2(n-1) n=101 So the sum would be (101(1+201))/2 10,201
For detailed explanations on arithmetic sequences and the formulas check out this website:
http://www.regentsprep.org/Regents/math/algtrig/ATP2/ArithSeq.htm
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