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Algebra Level 4

Find the number of terms with a non-zero coefficient in the expansion of: ${(1+x)}^{2018}+{(1+x^2)}^{2017}+{(1+x^3)}^{2016}$

1 solution

Aditya Kumar
Jan 15, 2016

Let P, Q and R be the terms of each of the expansions.

$n(P)=2019$ $\color{#3D99F6} {Simple\quad binomial\quad expansion}$
$n(Q)=2018$ $\color{#3D99F6} {Simple\quad binomial\quad expansion}$
$n(R)=2017$ $\color{#3D99F6} {Simple\quad binomial\quad expansion}$
$n(P,Q)=1010$ $\color{#3D99F6} {All\quad the\quad terms\quad in\quad Q\quad less\quad or\quad equal\quad to\quad 2018\quad are \quad in\quad P. \quad Then\quad N(P,Q)=1010 \quad because \quad exponents\quad in\quad Q \quad are\quad even)}$
$n(Q,R)=673$ $\color{#3D99F6} {All\quad the \quad terms\quad in\quad R\quad of \quad the\quad form \quad 6k\quad lessv or\quad equal\quad to\quad 4034\quad are\quad in\quad Q. \\\quad They\quad are \quad 0,6,12,18,...,4032.}$
$n(P,R)=673$ $\color{#3D99F6} {All\quad the\quad terms\quad in\quad R\quad to\quad 2018\quad are\quad in\quad P. \\\quad They\quad are\quad 0,3,6,...2016.}$
$n(P,Q,R)=337$ $\color{#3D99F6} {It\quad is\quad easy\quad to\quad check\quad that\quad we\quad are \quad talking\quad about\quad the\quad 6k\quad on\quad P\quad less\quad or\quad equal\quad\quad to\quad 2018: 0,6,12,...,2016. \quad That's\quad 336}$

Therefore number of dissimilar terms are: $n(P)+n(Q)+n(R)-n(P,Q)-n(Q,R)-n(R,P)+n(P,Q,R) \\=4035$