Question

Evaluate:


$\bf{\sum^6_{n=0\:}(3)^n}$



Need help A.S.A.P., thank you! :)

Answer 1

Answer:

1093

Explanation:

Given expression:

• $\sf \huge{ \sum _{n=0}^6\left(3\right)^n}$

Summation:

• $\sf a_0+\sum _{n=1}^63^n$

Formula:

• $\sf \sum\limits_{i=1}^n x_i = x_1 + x_2 + \dots + x_n$

Compute:

$\rightarrow \sf 3^0 + 3^1 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6$

$\rightarrow \sf 1 + 3 + 9 + 27 + 81 + 243 + 729$

$\rightarrow \sf 1093$

Answer 2

Answer:

$\boxed{\rm \: SUM = 1093 }$

Step-by-step explanation:

Given:

$\huge\rm {{ \sum}^6_{n=0\:} (3)^n}$

To Find:

Sum of the given finite series

Solution:

We'll use this formula:

$\boxed{\rm SUM = a \cdot\bigg( \cfrac{1 - r {}^{n} }{1 - r} \bigg)}$

where,

• a = first term
• r = ratio in between terms

Let's find out the ratio R by using this formulae:

$\rm \: r = \cfrac{a_{n + 1} }{a_n}$

According to the question,

• $\rm a_n = 3^n$
• $\rm a_{n+1}= 3^{n+1}$

Substitute:

$\rm \: r = \cfrac{3 {}^{n + 1} }{3 {}^{n} }$

Apply law of exponents:[a^m/a^n] = a^m-n

$\rm \: r = {3}^{n + 1 - n}$

Rearrange it as:

$\rm \: r = 3 {}^{n - n + 1}$

$\rm \: r = 3 {}^{1} = 3$

So,the ratio R is 3.

Now let's find out the First term A.

To find, substitute the value of n in 3^n:

• [It is given that n = 0]

$\rm \: a = 3 {}^{0}$

• [x^0 = 1]

$\rm \: a = 1$

Hence, first term A is 1.

NOW Substitute the value of the first term A and ratio R onto the formulae of sum:

$\rm \: a \cdot\bigg( \cfrac{1 - r {}^{n} }{1 - r} \bigg)$

• a = 1
• r = 3
• n = 7

Simplify.

$\rm SUM = \rm \: 1 \times \cfrac{1 - 3 {}^{7} }{ 1 - 1 \times 3}$

$\rm \: SUM = \cfrac{ - 2186}{1 - 3}$

$\rm \: SUM = \cfrac{ \cancel{ - 2186} \: \: {}^{1093} }{ \cancel{ - 2} \: \: ^{1} }$

$\rm \: SUM = 1093$

We're done!

Hence, the sum of the given Finite series is 1093.

$\rule{225pt}{2pt}$