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25 votes
25 votes
Evaluate:


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



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

User Costa Walcott
by
2.8k points

2 Answers

19 votes
19 votes

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 * \cfrac{1 - 3 {}^(7) }{ 1 - 1 * 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}

User Sachink
by
3.4k points
12 votes
12 votes

Answer:

1093

Step-by-step 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

User IanQ
by
2.6k points