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Sagot :
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]
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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