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Evaluate:


[tex]\bf{\sum^6_{n=0\:}(3)^n}[/tex]



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Answer:

1093

Explanation:

Given expression:

  • [tex]\sf \huge{ \sum _{n=0}^6\left(3\right)^n}[/tex]

Summation:

  • [tex]\sf a_0+\sum _{n=1}^63^n[/tex]

Formula:

  • [tex]\sf \sum\limits_{i=1}^n x_i = x_1 + x_2 + \dots + x_n[/tex]

Compute:

[tex]\rightarrow \sf 3^0 + 3^1 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6[/tex]

[tex]\rightarrow \sf 1 + 3 + 9 + 27 + 81 + 243 + 729[/tex]

[tex]\rightarrow \sf 1093[/tex]

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Answer:

[tex] \boxed{\rm \: SUM = 1093 }[/tex]

Step-by-step explanation:

Given:

[tex] \huge\rm {{ \sum}^6_{n=0\:} (3)^n}[/tex]

To Find:

Sum of the given finite series

Solution:

We'll use this formula:

[tex] \boxed{\rm SUM = a \cdot\bigg( \cfrac{1 - r {}^{n} }{1 - r} \bigg)}[/tex]

where,

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

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

[tex] \rm \: r = \cfrac{a_{n + 1} }{a_n}[/tex]

According to the question,

  • [tex]\rm a_n = 3^n[/tex]
  • [tex]\rm a_{n+1}= 3^{n+1}[/tex]

Substitute:

[tex] \rm \: r = \cfrac{3 {}^{n + 1} }{3 {}^{n} } [/tex]

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

[tex] \rm \: r = {3}^{n + 1 - n} [/tex]

Rearrange it as:

[tex] \rm \: r = 3 {}^{n - n + 1} [/tex]

[tex] \rm \: r = 3 {}^{1} = 3[/tex]

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]

[tex] \rm \: a = 3 {}^{0} [/tex]

  • [x^0 = 1]

[tex] \rm \: a = 1[/tex]

Hence, first term A is 1.

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

[tex] \rm \: a \cdot\bigg( \cfrac{1 - r {}^{n} }{1 - r} \bigg)[/tex]

  • a = 1
  • r = 3
  • n = 7

Simplify.

[tex] \rm SUM = \rm \: 1 \times \cfrac{1 - 3 {}^{7} }{ 1 - 1 \times 3} [/tex]

[tex] \rm \: SUM = \cfrac{ - 2186}{1 - 3} [/tex]

[tex] \rm \: SUM = \cfrac{ \cancel{ - 2186} \: \: {}^{1093} }{ \cancel{ - 2} \: \: ^{1} } [/tex]

[tex] \rm \: SUM = 1093[/tex]

We're done!

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

[tex] \rule{225pt}{2pt}[/tex]

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