Relationship

Algebra Level 1

( x n + x n − 1 + x n − 2 + … + x 3 + x 2 + x 1 + x 0 ) ( x − 1 ) = ? \large \displaystyle (x^{n}+x^{n-1}+x^{n-2} + \ldots + x^{3} + x^{2} + x^{1} + x^{0})(x-1) = \ ?

x − n x^{-n} x n − ( 1 / 2 n ) x^{n-(1/2n)} x 7 x^{7} x ( n 2 ) x{(n^{2})} x n + 1 − 1 x^{n+1} - 1 x n x{n}

Bryan Lee Shi Yang by Bryan Lee Shi Yang , 17, Malaysia

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1 solution

Expand: x n ( x − 1 ) + x n − 1 ( x − 1 ) + x n − 2 ( x − 1 ) + ⋯ + x 0 ( x − 1 ) = x n + 1 − x n + x n − x n − 1 + x n − 1 − ⋯ − x 1 + x 1 − 1 x^n(x-1)+x^{n-1}(x-1)+x^{n-2}(x-1)+\dots+x^0(x-1)\\=x^{n+1}-x^n+x^n-x^{n-1}+x^{n-1}-\dots-x^1+x^1-1 All the intermediate terms get cancelled and we are left with x n + 1 − 1 \boxed{x^{n+1}-1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nicely done.

Bonus question : What would the answer be if the expression is

( x n − x n − 1 + x n − 2 − … ± 1 ) ( x + 1 ) \large \displaystyle (x^{n}-x^{n-1}+x^{n-2} - \ldots \pm 1)(x+1)

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