n n
"Suppose ∑ i = 21. and. ∑ i^3 = 441"
i=1 i=1
i) Evaluate
n
Sn = ∑ [4i + (1/3) i^3)]
i=1
ii) Determine the value of n."
I got n = 6 / n = -7 using the "(n^2(n-1)^2)" formula but I'm not sure if it's right/how to do the first part. I'm very unsure of why "i = 21" and why "i^3 can be equal to 441".
n
∑ i = n(n+1)/2, so if ∑i=21, n=6
i=1
1^3+2^3+3^3+4^3+5^3+6^3 = 1+8+27+64+125+216 = 441
∑ [4i + (1/3) i^3)] = 4∑i + 1/3 ∑ i^3 = 4*21 + 1/3 * 441 = 84+147 = 231
To solve the given problem, let's break it down step by step.
i) Evaluate Sn = ∑[4i + (1/3)i^3], where i ranges from 1 to n.
To evaluate this sum, we need to substitute the values of i into the expression and add them up. Let's simplify the expression first:
Sn = 4∑i + (1/3)∑i^3
We know that ∑i = n(n+1)/2, which represents the sum of numbers from 1 to n.
Substituting this in, we have:
Sn = 4(n(n+1)/2) + (1/3)∑i^3
Now, let's focus on the sum of the cubes, ∑i^3.
ii) Determine the value of n.
We are given that ∑i = 21 and ∑i^3 = 441. Let's use this information to find the value of n.
We know that ∑i = n(n+1)/2, so we can set up the following equation:
n(n+1)/2 = 21
To solve for n, we can simplify the equation:
n^2 + n = 42
Rearranging the equation:
n^2 + n - 42 = 0
Now, we can factor this quadratic equation:
(n+7)(n-6) = 0
This gives us two possible solutions: n = -7 or n = 6.
However, since n represents the number of terms in the sum, we can discard the negative value (-7) as the number of terms cannot be negative.
Therefore, the value of n is 6.
To check if your calculation for n is correct, you can substitute n=6 back into the equation ∑i^3 = 441:
∑i^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3
Calculating this sum, we find:
∑i^3 = 1 + 8 + 27 + 64 + 125 + 216 = 441
The sum matches the given value of ∑i^3, confirming that n = 6 is the correct solution.
To recap:
i) Evaluate Sn = ∑[4i + (1/3)i^3], where i ranges from 1 to n.
ii) Determine the value of n using ∑i = 21 and ∑i^3 = 441.