What is the idea to sum this series of powers?∑n=1+∞(n2+n+1)xn−1

A small trick for this kind of summation∑n=1+∞(n2+n+1)xn−1n2=n(n−1)+n⇒n2+n+1=n(n−1)+2n+1∑n=1+∞(n2+n+1)xn−1=∑n=1+∞n(n−1)xn−1+2∑n=1+∞nxn−1+∑n=1+∞xn−1∑n=1+∞(n2+n+1)xn−1=x∑n=1+∞n(n−1)xn−2+2∑n=1+∞nxn−1+1x∑n=1+∞xn∑n=1+∞(n2+n+1)xn−1=x[∑n=1+∞xn]″+2[∑n=1+∞xn]′+1x[∑n=1+∞xn]

Best solutions to - "What is the idea to sum this series..."

simnelsbl4

Answered question

2022-02-28

What is the idea to sum this series of powers?

\sum_{n = 1}^{+ \infty} (n^{2} + n + 1) x^{n - 1}

Answer & Explanation

Radich606

Beginner2022-03-01Added 6 answers

A small trick for this kind of summation

\sum_{n = 1}^{+ \infty} (n^{2} + n + 1) x^{n - 1}

n^{2} = n (n - 1) + n \Rightarrow n^{2} + n + 1 = n (n - 1) + 2 n + 1

\sum_{n = 1}^{+ \infty} (n^{2} + n + 1) x^{n - 1} = \sum_{n = 1}^{+ \infty} n (n - 1) x^{n - 1} + 2 \sum_{n = 1}^{+ \infty} n x^{n - 1} + \sum_{n = 1}^{+ \infty} x^{n - 1}

\sum_{n = 1}^{+ \infty} (n^{2} + n + 1) x^{n - 1} = x \sum_{n = 1}^{+ \infty} n (n - 1) x^{n - 2} + 2 \sum_{n = 1}^{+ \infty} n x^{n - 1} + \frac{1}{x} \sum_{n = 1}^{+ \infty} x^{n}

\sum_{n = 1}^{+ \infty} (n^{2} + n + 1) x^{n - 1} = x [\sum_{n = 1}^{+ \infty} x^{n}]^{″} + 2 {[\sum_{n = 1}^{+ \infty} x^{n}]}^{'} + \frac{1}{x} [\sum_{n = 1}^{+ \infty} x^{n}]

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