Find the positive integer value of n such that
n^3−3/n^3+n^3−4/n^3+n^3−5/n^3+…+4/n^3+3/n^3=169.
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.
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
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
To find the positive integer value of n in the given equation:
n^3 - 3/n^3 + n^3 - 4/n^3 + n^3 - 5/n^3 + ... + 4/n^3 + 3/n^3 = 169
First, observe that each term in the given equation consists of n^3 and a fraction. The fractions are obtained by subtracting 3, 4, 5, ..., 4, 3 from n^3 successively.
Let's simplify the expression by combining like terms:
n^3 + n^3 + n^3 + ... + n^3 = 169
The number of terms in the simplified equation is equal to the sum of the positive integers from 3 to 4, which is:
Sum = (4 + 3)(4 - 3 + 1)/2 = (7)(2)/2 = 7
So, we have 7 terms of n^3 in the equation:
7n^3 = 169
Now, divide both sides of the equation by 7:
n^3 = 169/7
Since we are looking for a positive integer value for n, we need to find a cube of a number that is closest to 169/7. By trial and error, we can observe that n = 4 is the solution.
Therefore, the positive integer value of n is 4.