Answer

Let

f(n) = n^5+5n^4+10n^3+10n^2+5n+1

then, f(-1) = 0

n+1 is one of factors of f(n)

Group the terms to factor out n+1

f(n) = (n^4+n^4)+(4n^4+4n^3)+(6n^3+6n^2)+(4n^2+4n)+(n+1)

=(n+1)(n^4+4n^3+6n^2+4n+1)

Let g(n) = n^4+4n^3+6n^2+4n+1

then, g(-1) = 0

n+1 is one of factors of g(n)

Group the terms to factor out n+1 from g(n)

g(n) = n^4+4n^3+6n^2+4n+1

=(n^4+n^3)+(3n^3+3n^2)+(3n^2+3n)+(n+1)

=(n+1)(n^3+3n^2+3n+1)

=(n+1)(n+1)^3

=(n+1)^4

Therefore

f(n) = n^5+5n^4+10n^3+10n^2+5n+1

=(n+1)^5

Therefore

69^5+5\times 69^4+10\times 69^3+10\times 69^2+5\times69+1

=f(69) =(69+1)^5

=70^5

Steven Zheng posted 1 year ago

Scroll to Top