Given:
[tex](\frac{1}{4})+(\frac{1}{4})^2+(\frac{1}{4})^3+...+(\frac{1}{4})^5[/tex][tex]\sum_{j\mathop{=}1}^65(-3)^j[/tex]Required:
We need to compute the given sum.
Explanation:
Consider the given sum.
[tex](\frac{1}{4})+(\frac{1}{4})^2+(\frac{1}{4})^3+...+(\frac{1}{4})^5=(\frac{1}{4})+(\frac{1}{4})^2+(\frac{1}{4})^3+(\frac{1}{4})^4+(\frac{1}{4})^5[/tex][tex]Use\text{ \lparen}\frac{1}{4})^n=\frac{1}{4^n}.[/tex][tex]=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\frac{1}{4^5}[/tex][tex]The\text{ LCM is }4^5,\text{ make the denominator }4^5.[/tex][tex]=\frac{1\times4^4}{4\times4^4}+\frac{1\times4^3}{4^2\times4^3}+\frac{1\times4^2}{4^3\times4^2}+\frac{1\times4}{4^4\times4}+\frac{1}{4^5}[/tex][tex]=\frac{4^4}{4^5}+\frac{4^3}{4^5}+\frac{4^2}{4^5}+\frac{4}{4^5}+\frac{1}{4^5}[/tex][tex]=\frac{4^4+4^3+4^2+4+1}{4^5}[/tex][tex]=\frac{341}{1024}[/tex][tex](\frac{1}{4})+(\frac{1}{4})^2+(\frac{1}{4})^3+...+(\frac{1}{4})^5=\frac{341}{1024}[/tex]Consider the given expression.
[tex]\sum_{j\mathop{=}1}^65(-3)^j[/tex]Expand the sum.
[tex]\sum_{j\mathop{=}1}^65(-3)^j=5(-3)^1+5(-3)^2+5(-3)^3+5(-3)^4+5(-3)^5+5(-3)^6[/tex][tex]=5(-3)+5(9)+5(-27)+5(81)+5(-243)+5(729)[/tex][tex]=-15+45-135+405-1215+3645[/tex][tex]=-15-135-1215+45+405+3645[/tex][tex]=2730[/tex][tex]\sum_{j\mathop{=}1}^65(-3)^j=2730[/tex]Final answer:
[tex](\frac{1}{4})+(\frac{1}{4})^2+(\frac{1}{4})^3+...+(\frac{1}{4})^5=\frac{341}{1024}[/tex][tex]\sum_{j\mathop{=}1}^65(-3)^j=2730[/tex]