n n

Question

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".

Steve · Answer

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