Find the sum of the series
2 + 9/2! + 27/3! + 81/4! . . .
2 + 9/2! + 27/3! + 81/4! + ...
Sum from k = 0 to infinity of 3^k/k! -2
= exp(3) - 2
-1 + ( 3 + 3^2/2! + 3^3/3! + 3^4/4! ..)
but e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! ....
so
e^3 - 1 = 3 + 3^2/2! + 3^3/3! + 3^4/4! ....
so we have
-1 + e^3 -1 = -2 + e^3
i don't understand the last step. how did you get to -1 + e^3 -1 = -2 + e^3
To find the sum of the given series, we need to determine the common pattern or relationship between the terms.
Looking at the terms, we can see that each term is formed by raising the base number (2, 9, 27, 81) to the power of the term number, and dividing it by the factorial of the term number.
We can express each term as follows:
Term 1: 2^1 / 1!
Term 2: 3^2 / 2!
Term 3: 3^3 / 3!
Term 4: 3^4 / 4!
From this observation, we can generalize the nth term of the series as:
Term n: (n+1)^n / n!
Now, we can proceed to calculate the sum of the series.
Let Sn denote the sum of the first n terms:
Sn = Term 1 + Term 2 + Term 3 + ... + Term n
Substituting the general formula for the nth term:
Sn = (n+1)^n / n! + (n+2)^n / (n+1)! + (n+3)^n / (n+2)! + ... + (n+n)^n / (n+n-1)!
Now, let's simplify the expression to find a pattern.
Sn = [(n+1)^n / n!] + [(n+2)^n / (n+1)! ] + [(n+3)^n / (n+2)!] + ... + [(n+n)^n / (n+n-1) !]
Taking common denominator for all terms:
Sn = [(n+1)^n x (n+n-1)! + (n+2)^n x (n+n-2)! + (n+3)^n x (n+n-3)! + ... + (n+n)^n x n!] / [(n+n-1)! x (n+n-2)! x (n+n-3)! ... x n!]
Now, we can see that each term has a common factor of (n+n-1)!, that can be taken outside the summation sign:
Sn = (n+n-1)! x [ ( (n+1)^n / n! ) + ( (n+2)^n / (n+1)! ) + ( (n+3)^n / (n+2)! ) + ... + ( (n+n)^n / n! ) ]
Simplifying further:
Sn = (n+n-1)! x [ ( (n+1)^n + (n+2)^n + (n+3)^n + ... + (n+n)^n ) / n! ]
We can simplify the numerator by factoring out n^n:
Sn = (n+n-1)! x [ n^n x ( (1 + (n+1)^(-n)) + (1 + (n+2)^(-n)) + (1 + (n+3)^(-n)) + ... + (1 + (n+n)^(-n)) ) / n! ]
Now, let's observe the term inside the brackets:
( (1 + (n+1)^(-n)) + (1 + (n+2)^(-n)) + (1 + (n+3)^(-n)) + ... + (1 + (n+n)^(-n)) )
We can recognize this term as a sum of n+1 terms, where each term is of the form (1 + m^(-n)), where m takes values from (n+1) to (n+n).
However, finding a general formula for this sum is quite difficult.
Therefore, at this point, we can conclude that the expression for Sn cannot be further simplified or calculated using a direct formula.
To find the sum, we would need to know the values of 'n' and evaluate the expression for Sn using a calculator or computer program.