Question #312167

Find the limit of Xn=1/2[Xn-1+Xn]





1
Expert's answer
2022-03-19T02:41:10-0400

xn=12(Xn1+xn)xn=\frac{1}{2}(Xn-1 +xn)

limxn=x1+limk=1n(xk+1xk)\lim xn=x1+lim \sum_{k=1}^n(x_{k+1}-x_k)

=x1+x2x1+limk=2n(1)k1c2k1=x_1+x_2-x_1+lim \sum_{k=2}^n\frac{(-1)^{k-1}c}{2^{k-1}}





=x2+limk=1n2c22klimk=1n2c22k1=x_2+ lim \sum_{k=1}^{\frac{n}{2}}\frac{c}{2^{2k}}-lim \sum_{k=1}^{\frac{n}{2}}\frac{c}{2^{2k-1}}

=x2+clim(k=1n214kk=1n212122k1)=x_2+c lim(\sum_{k=1}^{\frac{n}{2}}\frac{1}{4^k}-\sum_{k=1}^{\frac{n}{2}}\frac{1}{2}\frac{1}{2^{2k-1}})

=x2+c(14(k=0n121214k)=x_2+c(\frac{1}{4}- (\sum_{k=0}^{\frac{n-1}{2}}\frac{1}{2}\frac{1}{4^k})





=x2+c(1423)=x2+23c=x_2+c(\frac{1}{4}-\frac{2}{3}) =x_2+\frac{2}{3}c

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