About high school mathematics sequence sum, how to do? Given an = 2 to the nth power, BN = 2n Find TN = B1 / A1 + B2 / A2 + & ﹥ 8226; & ﹥ 8226; & ﹥ 8226; + BN / an (B is the numerator, a is the denominator)

About high school mathematics sequence sum, how to do? Given an = 2 to the nth power, BN = 2n Find TN = B1 / A1 + B2 / A2 + & ﹥ 8226; & ﹥ 8226; & ﹥ 8226; + BN / an (B is the numerator, a is the denominator)

That's right. Subtraction by dislocation
Tn=2/2+(2*2)/(2^2)+(2*3)/(2^3)+...+2n/(2^n)------①
Tn/2=2/(2^2)+(2*2)/(2^3)+(2*3)/(2^4)+...+2n/(2^(n+1))-------②
① - 2
Tn/2=1+2/(2^2)+2/(2^3)+...+2/(2^n)-2n/(2^(n+1))
=1+2[1/(2^2)+1/(2^3)+...+1/(2^n)]-2n/(2^(n+1))
=1+2[2^(-2)+2^(-3)+...+2^(-n)]-2n/(2^(n+1))
=1+2[(1-(1/2)^n)-1/2]-2n/(2^(n+1))
=1+ 2(1-(1/2)^n)-1-2n/(2^(n+1))
= 2(1-(1/2)^n)-2n/(2^(n+1))
So TN = 2 [2 (1 - (1 / 2) ^ n) - 2n / (2 ^ (n + 1))]
=4-(1/2)^(n-2)-4n/(2^(n+1))
=4-(1/2)^(n-2)-n/(2^(n-1))
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