Let [ x ] denote greatest integer less than or equal to x.

Question

Let [ x ] denote greatest integer less than or equal to x. If for n$$\in$$N,

$${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $$,

then $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1} $$ is equal to :

Accepted Answer

$${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $$

$$(1 - x + {x^3}) = {a_0} + {a_1}x + {a_2}{x^2} + ...... + {a_{3n}}{x^{3n}}$$

Put x = 1

$$1 = {a_0} + {a_1} + {a_2} + {a_3} + {a_4} + ........ + {a_{3n}}$$ ...... (1)

Put x = -1

$$1 = {a_0} - {a_1} + {a_2} - {a_3} + {a_4} + ........( - 1){}^{3n}{a_{3n}}$$ ..... (2)

Add (1) + (2)

$$ \Rightarrow {a_0} + {a_2} + {a_4} + {a_6} + ...... = 1$$

Sub (1) - (2)

$$ \Rightarrow {a_1} + {a_3} + {a_5} + {a_7} + ...... = 0$$

Now, $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}}}  + 4\sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j }}} + 1 $$

$$ = ({a_0} + {a_2} + {a_4} + ......) + 4({a_1} + {a_3} + .....)$$

$$ = 1 + 4 \times 0$$ + 1

$= 1 + 1 = 2$

Solutions · 1