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### Section 4-3 : Series - Basics

For problems 1 – 4 perform an index shift so that the series starts at $$n = 4$$.

1. $$\displaystyle \sum\limits_{n = 8}^\infty {\frac{{2 + n}}{{5 - n}}}$$
2. $$\displaystyle \sum\limits_{n = 2}^\infty {\frac{{{3^{n + 2}}}}{{1 - {4^{3n + 1}}}}}$$
3. $$\displaystyle \sum\limits_{n = 0}^\infty {{{\left( { - 2} \right)}^{2 - n}}\,{{\bf{e}}^{3n}}}$$
4. $$\displaystyle \sum\limits_{n = 5}^\infty {\frac{{{{\left( { - 1} \right)}^{3 + n}}{n^2}}}{{{n^2} - 2n + 1}}}$$
5. Strip out the first 4 terms from the series $$\displaystyle \sum\limits_{n = 0}^\infty {{3^n}\,{6^{2 - n}}}$$.
6. Strip out the first 2 terms from the series $$\displaystyle \sum\limits_{n = 3}^\infty {\frac{4}{{{n^2} + n + 1}}}$$.
7. Given that $$\displaystyle \sum\limits_{n = 4}^\infty {n\,{4^{ - n}}} = 0.02257$$ determine the value of $$\displaystyle \sum\limits_{n = 1}^\infty {n\,{4^{ - n}}}$$.
8. Given that $$\displaystyle \sum\limits_{n = 3}^\infty {\frac{{n + 1}}{{{n^3}}}} = 0.47199$$ determine the value of $$\displaystyle \sum\limits_{n = 5}^\infty {\frac{{n + 1}}{{{n^3}}}}$$.