Sum (2-3 * 5 ^ - 1) + (4-3 * 5 ^ - 2) +... + (2n-3 * 5 ^ - n) As the title I don't know if I wrote the wrong question. Can the last n be equal to 71? If I wait, the answer is much better? Wrong, the last negative n can be changed to 71? ..

Sum (2-3 * 5 ^ - 1) + (4-3 * 5 ^ - 2) +... + (2n-3 * 5 ^ - n) As the title I don't know if I wrote the wrong question. Can the last n be equal to 71? If I wait, the answer is much better? Wrong, the last negative n can be changed to 71? ..

(2-3*5^-1)+(4-3*5^-2)+...+(2n-3*5^-n)
=2+4+6+8+…… +2n-3(5^-1+5^-2+5^-3+…… +5^-n)
=n^2+n-3*(1/5)*(1-(1/5)^n)/(1-1/5)
=n^2+n-3/4+3/4*(1/5)^n