Question Question B4: The ideas of linear recurrences show how some of the summation formulas from section 2.4 arise. Let $$a_{n}=\sum_{i=1}^{k}(n+2)(n+1)(n)$$, and observe that $$a_{n}=a_{n-1}+n^{3}+3 n^{2}+2 n$$ is a non-homogeneous linear recurrence with initial condition $$a_{0}=0$$. (a) Show that $$a_{n}$$ must be a polynomial function of $$n$$ with degree 4 . (b) Determine the coefficients of this polynomial. Check that this is the same formula you would obtain by using the table of summations from section 2.4?

Transcribed Image Text: Question B4: The ideas of linear recurrences show how some of the summation formulas from section 2.4 arise. Let $$a_{n}=\sum_{i=1}^{k}(n+2)(n+1)(n)$$, and observe that $$a_{n}=a_{n-1}+n^{3}+3 n^{2}+2 n$$ is a non-homogeneous linear recurrence with initial condition $$a_{0}=0$$. (a) Show that $$a_{n}$$ must be a polynomial function of $$n$$ with degree 4 . (b) Determine the coefficients of this polynomial. Check that this is the same formula you would obtain by using the table of summations from section 2.4?
Transcribed Image Text: Question B4: The ideas of linear recurrences show how some of the summation formulas from section 2.4 arise. Let $$a_{n}=\sum_{i=1}^{k}(n+2)(n+1)(n)$$, and observe that $$a_{n}=a_{n-1}+n^{3}+3 n^{2}+2 n$$ is a non-homogeneous linear recurrence with initial condition $$a_{0}=0$$. (a) Show that $$a_{n}$$ must be a polynomial function of $$n$$ with degree 4 . (b) Determine the coefficients of this polynomial. Check that this is the same formula you would obtain by using the table of summations from section 2.4?
&#12304;General guidance&#12305;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 $$\mathrm{{a}_{{{n}}}}$$&#160;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