Seemingly difficult but too easy- Part 3!!!

Number Theory Level 1

Let n n be a positive integer. Then, find the smallest possible value of 1 − n + n 2 − n 3 + n 4 . . . . . . . . . + n 1000 1-n+n^{2}-n^{3}+n^{4}.........+n^{1000} .


The answer is 1.

Yash Singhal by Yash Singhal , 22, India

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2 solutions

Sanjeet Raria
Sep 9, 2014

( x n + y n ) = ( x + y ) ( x n − x n − 1 y + x n − 2 y 2 − x n − 3 y 3 + . . . . . + y n ) (x^n+y^n)=(x+y)(x^n-x^{n-1}y+x^{n-2}y^2-x^{n-3}y^3+.....+y^n) Hence following the result, the given series is, ( 1 + n 1001 1 + n ) (\frac{1+n^{1001}}{1+n}) Now clearly, smallest value will be at n = 1 n=1 . Smallest value is 1 \boxed{1}

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well done nice solution

Mardokay Mosazghi - 6 years, 8 months ago

Nice solution! I assumed it as S S . Then calculated − n S -nS and subtrated them to get that. Btw you may like to precede parentheses by \color{#3D99F6}{\text{\Big}} to get bigger parentheses.

Pranjal Jain - 6 years, 6 months ago
James Moors
Oct 1, 2014

For n > 1 , n 1000 n>1, n^{1000} will dominate the sum. n ∈ N n\in\mathbb{N} so n=1

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