Find the positive integer value of n such that

Question

n^3−3/n^3+n^3−4/n^3+n^3−5/n^3+…+4/n^3+3/n^3=169.

Answer 1

I have a strong feeling that you mean:

(n^3−3)/n^3 + (n^3−4)/n^3 + (n^3−5)/n^3+…+ 4/n^3 + 3/n^3=169

please confirm before I attempt it.

Answer 2

Yes Reiny, the question is [(n^3−3)/n^3] + [(n^3−4)/n^3] + [(n^3−5)/n^3] + … + [4/n^3] + [3/n^3] = 169

Answer 3

ok then,

(n^3−3)/n^3 + (n^3−4)/n^3 + (n^3−5)/n^3+…+ 4/n^3 + 3/n^3=169

(n^3−3) + (n^3−4) + (n^3−5)+…+ 4 + 3 =169n^3

The left side is an arithmetic series where
a = n^3 - 3 , and d = -1
we don't know how many terms, let the number of terms be N
using term(N) = a + (N-1)d
3 = n^3 - 3 + (N-1)(-1)
6 = n^3 - N + 1
N = n^3 - 5

sum of the left side
= (n^3-5)/2 [ 2(n^3 - 3) + (n^3 - 6)(-1) ]
= (n^3 - 5)/2 [ 2n^3 - 6 - n^3 + 6 ]
= (n^3 - 5)/2 [ n^3]

so (n^3-5)(n^3)/2 = 169n^3
n^3 - 5 = 338
n^3 = 343
n = 7 and N = 338 terms

check:
If n = 7
LS = (343-3)/343 + (343-5)/343 + ... + 3/343
= 340/343 + 339/343 + ... + 3/343

sum of 338 terms
= (338/2) (first + last)
= 169(340/343 + 3/343)
= 169(343/343) = 169 = RS

my answer is correct

there are 338 terms, and n = 7