First slide

Binomial theorem for positive integral Index

Question

Let $E = 1^{2017} + 2^{2017} + 3^{2017} + \dots + 2016^{2017}$ then E is divisible by

Moderate

A

2016
B

2017
C

4033
D

2016²

Solution

For $1 \leq x \leq 1013$
$\begin{array}{l} x^{2017} + (2017 - x)^{2017} \\ = x^{2017} +^{2017} C_{0} (2017)^{2017} +^{2017} C_{1} (2017)^{2016} (- x) +^{2017} C_{2} (2017)^{2015} (- x)^{2} + \dots +^{2017} C_{2016} (2017) (- x)^{2016} + (- x)^{2017} \\ = (2017) m \end{array}$
Where $m =^{2017} C_{0} (2017)^{2016} +^{2017} C_{1} (2017)^{2015} (- x) + \dots +^{2017} C_{2016} (- x)^{2016}$
is a positive integer.
Now, write E as sum of 1013 groups as follows:
$E = (1^{2017} + 2016^{2017}) + (2^{2017} + 2015^{2017}) + \dots + (1013^{2017} + 1014^{2017})$
and note that each group is divisible by 2017

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