This question is similar to the last one that I did but this have been change completely. What is the pattern to this...
" For the first year is 2000$
For the second year is (2000(0.075) +2000) +2000 = 4150$
For the third year is (2000(0.075)(2) +2000) + (2000(0.075) +2000) + 2000 = 6450$
For the third year is (2000(0.75)(3) +2000)+(2000(0.075)(2) +2000) + (2000(0.075) +2000) + 2000 = 12950$
What is the nth year?"
Easy Math Editor
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The sum is 2000+m=1∑n2000(1+n(0.075)).
Apply the algebraic identity 1+2+⋯+n=2n(n+1).
Can you take it from here?