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Question

Question: How do you find the sum of \[\sum {\dfrac{{{3^n} + {4^n}}}{{{5^n}}}} \] from \[n\] is \[[0,\infty )\...

How do you find the sum of 3n+4n5n\sum {\dfrac{{{3^n} + {4^n}}}{{{5^n}}}} from nn is [0,)[0,\infty )?

Explanation

Solution

In the question involving the concept of the geometric series, it is important to note that the sum of the terms in an infinite G.P. is given by, a1r\dfrac{a}{{1 - r}} where a is the first term and r is the common ratio. Also, the sum of the terms of any infinite G.P. is given bya+ar+ar2+ar3+.........a + ar + a{r^2} + a{r^3} + .......... Thus, using the given conditions and applying the formula of sum of infinite G.P., you will get the required answer.

Complete step by step solution:
We are given to find sum of terms 3n+4n5n\sum {\dfrac{{{3^n} + {4^n}}}{{{5^n}}}} from nn is [0,)[0,\infty )
Let us simplify the obtained expression to get a clear picture of this condition. 3n+4n5n=(35)n+(45)n\sum {\dfrac{{{3^n} + {4^n}}}{{{5^n}}}} = \sum {{{\left( {\dfrac{3}{5}} \right)}^n} + \sum {{{\left( {\dfrac{4}{5}} \right)}^n}} } …..(1)
n=0(35)n=1+(35)1+(35)2+........\Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{3}{5}} \right)}^n} = 1 + } {\left( {\dfrac{3}{5}} \right)^1} + {\left( {\dfrac{3}{5}} \right)^2} + ........\infty
The first term and common ratio of the above series (on the right side first term in equation (1) )
a=1,r=(35)\Rightarrow a = 1,r = \left( {\dfrac{3}{5}} \right)
Also, we know that the sum of the terms of the infinite series is given by,
Sum=a1r=1135=52Sum = \dfrac{a}{{1 - r}} = \dfrac{1}{{1 - \dfrac{3}{5}}} = \dfrac{5}{2}
Again we repeat the same process for the other term in equation (1)
n=0(45)n=1+(45)1+(45)2+........\Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{4}{5}} \right)}^n} = 1 + } {\left( {\dfrac{4}{5}} \right)^1} + {\left( {\dfrac{4}{5}} \right)^2} + ........\infty
The first term and common ratio of the above series
a=1,r=(45)\Rightarrow a = 1,r = \left( {\dfrac{4}{5}} \right)
Also, we know that the sum of the terms of the infinite series is given by,
Sum=11(45)=5Sum = \dfrac{1}{{1 - \left( {\dfrac{4}{5}} \right)}} = 5
Now we write from equation (1)
Value of 3n+4n5n\sum {\dfrac{{{3^n} + {4^n}}}{{{5^n}}}} =(35)n+(45)n= \sum {{{\left( {\dfrac{3}{5}} \right)}^n} + \sum {{{\left( {\dfrac{4}{5}} \right)}^n}} } =5+52=152=7.5 = 5 + \dfrac{5}{2} = \dfrac{{15}}{2} = 7.5
Therefore, the value of 3n+4n5n\sum {\dfrac{{{3^n} + {4^n}}}{{{5^n}}}} where nn is [0,)[0,\infty ) will be 7.57.5

Note:
In order to get command over these types of questions, you need to have a grip over the formulas used in the concept of G.P. Also, you need to be familiar with how to calculate the common ratio. Knowing the formula and applying the given conditions to form the equations will lead you to your final answer. Don’t confuse yourself between the common ratios of AP and GP, as one thing being wrong will make your entire solution wrong.