This one

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?"

Note by A Former Brilliant Member
3 years, 7 months ago

No vote yet
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Comments

The sum is 2000+m=1n2000(1+n(0.075)) \displaystyle 2000 + \sum_{m=1}^n 2000(1 + n(0.075)) .

Apply the algebraic identity 1+2++n=n(n+1)21 + 2 + \cdots + n = \dfrac{n(n+1)}2 .

Can you take it from here?

Pi Han Goh - 3 years, 7 months ago
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