Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2(a). To prove that \( \mathrm{{a}_{{{n}}}} \)is a polynomial function of n with degree 4, we will show that the recurrence relation implies that \( \mathrm{{a}_{{{n}}}} \) can be written as a polynomial of degree 4 in terms of n. We can start by expanding \( \mathrm{{a}_{{{n}-{1}}}} \)in the recurrence relation:\( \mathrm{{a}_{{{n}-{1}}}={a}_{{{n}-{2}}}+{\left({n}-{1}\right)}^{{3}}+{3}{\left({n}-{1}\right)}^{{2}}+{2}{\left({n}-{1}\right)}} \) Substituting this into the recurrence relation gives:\( \mathrm{{a}_{{{n}}}={a}_{{{n}-{2}}}+{\left({n}-{1}\right)}^{{3}}+{3}{\left({n}-{1}\right)}^{{2}}+{2}{\left({n}-{1}\right)}+{n}^{{3}}+{3}{n}^{{2}}+{2}{n}} \) Expanding \( \mathrm{{a}_{{{n}-{2}}}} \)in this equation and using the initial condition \( \mathrm{{a}_{{{0}}}={0}} \), we get:\( \mathrm{{a}_{{{n}}}={n}^{{3}}+{3}{n}^{{2}}+{2}{n}+{\sum_{{{i}={1}}}^{{{n}-{1}}}}{\left({\left({i}+{1}\right)}^{{3}}+{3}{\left({i}+{1}\right)}^{{2}}+\right.}} \) \( \mathrm{{2}{\left({i}+{1}\right)}{)}} \) Simplifying this expression, we get:\( \mathrm{{a}_{{{n}}}={n}^{{3}}+{3}{n}^{{2}}+{2}{n}+{\sum_{{{i}={2}}}^{{{n}}}}{\left({i}^{{3}}+{3}{i}^{{2}}+{2}{i}+{3}\right)}} \) We can now evaluate the sum using the formulas from section 2.4:\( \mathrm{{\sum_{{{i}={1}}}^{{{n}}}}{i}^{{3}}={\left[{\frac{{{n}{\left({n}+{1}\right)}}}{{{2}}}}\right]}^{{2}}} \)\( \mathrm{{\sum_{{{i}={1}}}^{{{n}}}}{i}^{{2}}={\frac{{{n}{\left({n}+{1}\right)}{\left({2}{n}+{1}\right)}}}{{{6}}}}} \)\( \mathrm{{\sum_{{{i}={1}}}^{{{n}}}}{i}={\frac{{{n}{\left({n}+{1}\right)}}}{{{2}}}}} \) Substituting these formulas into the expression for ... See the full answer