Rewrite the following sequence defined by a Recurrence Relation with the Sigma notation.

Question

The Sequence <$c_n$>$_{n∈N}$ is defined with the following Recurrence Relation: $$ c_o = 1$$ $$ c_n = c_{n-1} + 3^n <\forall n ∈ N^*>$$

Rewrite the definition of this sequence in sigma notation.

The first 5 terms are : $$ (0+3^0), (1+3^1), (4+3^2), (13+3^3), (40+3^4), ... ,(c_{n-1} + 3^n) $$ $$1,4,13,40,121, ... ,(c_{n-1} + 3^n)$$

Now to rewrite it in sigma notation... $$\sum_{i=0}^{n} ??$$

That's as far as I could go... any hints or help will be well appreciated.

Answer 1

You have

\begin{align} c_0 & =1& =\sum_{k=0}^0 3^k\\ c_1&= 1+3 & =\sum_{k=0}^1 3^k\\ c_2 &= 1+3+3^2& =\sum_{k=0}^2 3^k \end{align}

That is:

$$c_n=\sum_{k=0}^n 3^k.$$

Answer 2

From $$ c_n - c_{n-1} =3^n,\qquad n=1,2,\cdots, $$ by summing terms telescope on the left hand side giving $$ c_n-c_0=\sum_{k=1}^n3^k $$ or, since $c_0=1$,

$$ c_n=\sum_{k=0}^n3^k. $$