Question

Rewrite the given expression as a sum whose generic term involves \(x^{n} .\) $$ x \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}+\sum_{n=0}^{\infty} a_{n} x^{n} $$

Short answer

Question: Express the following as a sum whose generic term involves \(x^n\): $$ x \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}+\sum_{n=0}^{\infty} a_n x^n $$ Answer: $$ S(x) = \sum_{n=0}^{\infty} a_n [(n-1)n x^{n} + x^{n}] $$

Step-by-step solution

Observe the given expressions

Notice that the first sum has terms with exponent \(x^{n-2}\) and coefficients \(n(n-1)a_n\), and the second sum has exponent \(x^n\) and coefficients \(a_n\). Our goal is to combine these expressions into a single power series.

Combine the expressions

We need to ensure that the exponents of both summations match so that we can combine these series. In the first summation, we substitute \((n-2)\) with \(m\) (where \(m = n-2\)). Then \(n=m+2\). Rewrite the first summation with the appropriate substitutions: $$ x \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2} = x \sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^{m} $$ Now both summations are in terms of \(x^m\).

Combine both sums

Rewrite the expression combining both sums: $$ S(x) = x \sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^{m}+\sum_{m=0}^{\infty} a_{m} x^{m} $$

Factor out the common \(x^m\) term

Factor out the common \(x^m\) term and rewrite the expression as one summation: $$ S(x) = \sum_{m=0}^{\infty} [(m+2)(m+1) a_{m+2} x^{m+1} + a_m x^{m}] $$

Rewrite as a generic term \(x^n\)

Since the exponent \(m+1\) in the first part of the summation ranges from \(1\) to \(\infty\), we can rewrite the series with \(n\) as the exponent: $$ S(x) = \sum_{n=0}^{\infty} [(n-1)n a_n x^{n} + a_n x^{n}] $$

Group the coefficients of \(x^n\)

To achieve the final form, we need to group the coefficients of \(x^n\). Combine terms with \(a_n\): $$ S(x) = \sum_{n=0}^{\infty} a_n [(n-1)n x^{n} + x^{n}] $$ Therefore, the expression has been rewritten as a sum whose generic term involves \(x^n\).