Question

What is the least value of $n$ such that $\left( {1 + 3 + {3^2} + \cdots + {3^n}} \right)$ exceeds $2000$ ?
A) $7$
B) $5$
C) $8$
D) $6$

Answer 1

We have to first observe that the given expression expressed in the bracket is a geometric progression with starting term one. We will have to find the sum of the given geometric progression and compare it with the given limit of the integer.

Complete step-by-step answer:
The given expression is $\left( {1 + 3 + {3^2} + \cdots + {3^n}} \right)$.
Observe that the series is a geometric progression with the first term of the series being $1$ and the common ratio for the series being $r = 3$ .
We have to find the least value of $n$ such that the sum $\left( {1 + 3 + {3^2} + \cdots + {3^n}} \right)$ exceeds the number $2000$ .
For a geometric progression with first term $a$ and the common ratio $r$, the sum is given as follows:
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
In the given expression we have $a = 1,r = 3$ thus the sum is given by:
$\Rightarrow$ ${S_n} = \dfrac{{1\left( {{3^n} - 1} \right)}}{{3 - 1}}$
We have to find the value of $n$ such that the maximum value is $2000$ .
Therefore, we will put that as the sum.
Hence, we get the following:
$\Rightarrow$ $2000 = \dfrac{{{3^n} - 1}}{2}$
That means we get,
$\Rightarrow$ $4001 = {3^n}$
We know that ${3^7} > 2000$ .
Therefore, the least value of $n = 7$ .
Hence, the correct option is A.

Note: Notice that first we established the type of the series that is expressed in. later we found the sum of the given series so that we can express the sum in general form. After that we will use the general expression and the given sum limit to find the bound.