Question

Find the sum Sn below:


`\displaystyle \large{S = n \cdot 1 + (n - 1) \cdot 3 + (n - 2) \cdot {3}^(2) + (n - 3) \cdot {3}^(3) + ... + 2 \cdot {3}^(n - 2) + 1 \cdot {3}^(n - 1) }`
Please show your work too. Thanks!​

Answer 1

We can write S as

`\displaystyle S = \sum_(k=0)^(n-1) (n-k)3^k`

and expand it as

`\displaystyle S = n \sum_(k=0)^(n-1) 3^k - \sum_(k=0)^(n-1) k\cdot3^k`

The first sum is geometric, nothing tricky:

`\displaystyle\sum_(k=0)^(n-1) 3^k = 1 + 3 + 3^2 + \cdots + 3^(n-1) \\\\ \implies 3\sum_(k=0)^(n-1) 3^k = 3 + 3^2 + 3^3 + \cdots + 3^n \\\\ \implies -2\sum_(k=0)^(n-1) 3^k = 1 - 3^n \\\\ \implies \sum_(k=0)^(n-1) 3^k = \frac{3^n-1}2`

For the second sum, you can use the same method employed in another question of yours (24494877) to find

`\displaystyle \sum_(k=0)^(n-1) k\cdot 3^k = \frac{(2n-3)3^n+3}4`

So this sum comes out to

`\displaystyle S = n\cdot\frac{3^n-1}2 - \frac{(2n-3)3^n+3}4 \\\\ S = \frac{2n\cdot3^n-2n - (2n-3)3^n-3}4 \\\\ \boxed{S = \frac{3^(n+1)-2n-3}4}`