Can any one please expand
1 (n-1)n(n+1)(n+2) and
2 n(n+1)(n+2)(n+3)
Thanks - just the summary will do
1 (n-1)n(n+1)(n+2) and
2 n(n+1)(n+2)(n+3)
Thanks - just the summary will do
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1.
(n² - n) (n² + 3n + 2)
n^4 + 3n³ + 2n²
____- n³ - 3n² - 2n
n^4 + 2n³ - n² - 2n
2.
(n² + n)(n² + 5n + 6)
n^4 + 5n³ + 6n²
_____ n³ + 5n² + 6n
n^4 + 6n³ + 11n² + 6n
(n² - n) (n² + 3n + 2)
n^4 + 3n³ + 2n²
____- n³ - 3n² - 2n
n^4 + 2n³ - n² - 2n
2.
(n² + n)(n² + 5n + 6)
n^4 + 5n³ + 6n²
_____ n³ + 5n² + 6n
n^4 + 6n³ + 11n² + 6n
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1.
(n^2-n)(n^2+3n+2)
=n^4+3n^3+2n^2-n^3-3n^2-2n=n^4+2n^3-n^2…
2.
(n^2+n)(n^2+5n+6)
=(n^4+5n^3+6n^2+n^3+5n^2+6n)
=n^4+6n^3+11n^2+6n
(n^2-n)(n^2+3n+2)
=n^4+3n^3+2n^2-n^3-3n^2-2n=n^4+2n^3-n^2…
2.
(n^2+n)(n^2+5n+6)
=(n^4+5n^3+6n^2+n^3+5n^2+6n)
=n^4+6n^3+11n^2+6n
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1. (n-1)n(n+1)(n+2) = (n-1)n(n^2+3n+2) = (n-1)(n^3 + 3n^2 + 2n) = n^4 + 3n^3 + 2n^2 - (n^3 + 3n^2 + 2n) =
n^4 + 2n^3 - n^2 - 2n
2. n(n+1)(n+2)(n+3) = n(n+1)(n^2 + 5n + 6) = n(n^3 + 5n^2 + 6n + n^2 + 5n + 6) =
n^4 + 6n^3 + 11n^2 + 6n
n^4 + 2n^3 - n^2 - 2n
2. n(n+1)(n+2)(n+3) = n(n+1)(n^2 + 5n + 6) = n(n^3 + 5n^2 + 6n + n^2 + 5n + 6) =
n^4 + 6n^3 + 11n^2 + 6n