Algebraically find the sum

Question

50

∑((n-1/n!) + n³⁾

n=30

Answer 1

50 times sum of (n-1/n! +n^3) with n= 30

= 50 x sum (30 - 1/30! + 30^3)

= 50 x (30 + 27000 - 1(/2.6525286 x 10^32))

= nearly 50(27030 - 0)

= about 1,351,500

which is surprisingly close to what the other tutor came up with

but

it depends on what you intended as the problem. it's slightly ambiguous

Answer 2

Ravi

I have been working on this problem for several hours. I think I have a solution but it isn't pretty.

Begin by simplifying the fraction

n-1 = n - 1 = 1 - 1

n ! n ! n ! (n - 1)! n!

∑ n - 1 + n^{3 =} ∑₃₀⁵⁰ [ 1 - 1 + n³ ] = We must break up the n³ into two parts.

n! (n - 1)! n!

∑₃₀⁵⁰ (1 / (n-1)! - ∑₃₀⁵⁰ (1/n!) + ∑₁⁵⁰ n³ - ∑₁²⁹ n³ = Now expanding each summation.

Using the formula for ∑₁^k k³ = (1/4) (k²)(k + 1)² on the third and fourth terms...

(1/29! + 1/30! + 1/31! + ... 1/49!) - ( 1/30! + 1/31! + 1/32! + ... 1/50!) +

((1/4)(50²)(51²)) - ((1/4)(29²)(30²)

In the first two parts, the fraction will often cancel each other, leaving...

[(1/29!) - (1/50!)] + [(1/4)(2500)(2601) - (1/4)(841)(900)] =

[(1/29!) - (1/50!)] + (1625625 -189225) = [(1/29!) - (1/50!)] + 1,436,400

If this answer is incorrect, maybe it will give you a clue about the process I hope.

Linda

Answer 3

Calculate the sum of each term then combine.