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$$a_1=1$$ $$a_{n+1}=(n+3)a_n$$

How can I get to the answer of this, which is:

$$a_n=\frac{(n+2)!}{6}$$

4 Answers 4

4

$$a_{n+1}=(n+3)a_n=(n+3)(n+2)a_{n-1}=(n+3)(n+2)(n+1)a_{n-2}=...$$

can you find the pattern now ?

3

write it as follows $$\frac { { a }_{ n+1 } }{ { a }_{ n } } =n+3\\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } =4,\frac { { a }_{ 3 } }{ { { a } }_{ 2 } } =5,\frac { { a }_{ 4 } }{ { a }_{ 3 } } =6,...\frac { { a }_{ n } }{ { a }_{ n-1 } } =n+2\\ \\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } \cdot \frac { { a }_{ 3 } }{ { { a } }_{ 2 } } \cdot \frac { { a }_{ 4 } }{ { a }_{ 3 } } \cdot ...\cdot \frac { { a }_{ n } }{ { a }_{ n-1 } } =\frac { \left( n+2 \right) ! }{ 6 } $$

  • 0
    Thanks! although, — why divide by 6? – 2017-01-21
  • 0
    @Niki,because $4\cdot 5\cdot 6\cdot ...\cdot n\cdot \left( n+1 \right) \left( n+2 \right) =\frac { 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot ...\cdot n\cdot \left( n+1 \right) \left( n+2 \right) }{ 1\cdot 2\cdot 3 } =\frac { \left( n+2 \right) ! }{ 6 } $ – 2017-01-21
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Upon taking logarithms we have,

$$\ln a_{n+1}=\ln a_n+\ln (n+3)$$

Little Algebra and introducing dummy variable.

$$\ln a_{i+1}-\ln a_{i}=\ln (i+3)$$

Summing both sides from $i=1$ to $n-1$ and noticing that the sum is a telescoping sum,

$$\ln a_{n}-\ln 1=\sum_{i=1}^{n-1} \ln (i+3)$$

Logarithmic rules,

$$\ln a_{n}=\ln \prod_{i=1}^{n-1} (i+3)$$

$$a_n=\prod_{i=1}^{n-1} (i+3)$$

Note the $(1)(2)(3)$ we would like is missing.

$$3! a_n=(1)(2)(3)\prod_{i=1}^{n-1} (i+3)$$

Noting definition of factorial.

$$3! a_n=(n+2)!$$

Solving and simplifying,

$$a_n=\frac{(n+2)!}{6}$$

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Let ${\{a_n\}}_{n = 1}^{\infty}$ be the number sequence $$ a_1 = 1 \qquad \mbox{ and } \qquad a_{n + 1} = (n + 3) a_n \quad \mbox{ for all } \quad n = 1 , 2 , \ldots $$ and let ${\{b_n\}}_{n = 1}^{\infty}$ be the number sequence $$ b_n = \frac{(n + 2) !}{6} \quad \mbox{ for all } \quad n = 1 , 2 , \ldots\mbox{,} $$ so let's show that $a_n = b_n$ for all $n = 1 , 2 , \ldots$ and we do using induction method. On the one hand, $$ a_1 = 1 = \frac{6}{6} = \frac{(1 + 2) !}{6} = b_1\mbox{.} $$ On the other hand, let $n$ be a natural number, with $n > 1$, let's suppose that $a_n = b_n$ and let's show the equality $a_{n + 1} = b_{n + 1}$. It isn't difficult: $$ a_{n + 1} = (n + 3) a_n = (n + 3) b_n = (n + 3) \frac{(n + 2) !}{6} = \frac{(n + 3) !}{6} = \frac{((n + 1) + 2) !}{6} = b_{n + 1}\mbox{.} $$