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Composite Function

If 1+x+x2+x...

Question

If ${(1 + x + x^{2} + x^{3})}^{100}$ = $a_{0} + a_{1} x + a_{2} x^{2} + \cdot \cdot \cdot + a_{300} x^{300}$ then

A

a_{0} + a_{1} + a_{2} + a_{3} + \cdot \cdot \cdot + a_{300}

is divisible by 1024

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B

a_{0} + a_{2} + a_{4} + \cdot \cdot \cdot + a_{300} = a_{1} + a_{3} + \cdot \cdot \cdot + a_{299}

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C

coefficients equidistant from beginning and end are equal

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D

a_{1} = 100

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Solution

The correct options are
A $a_{0} + a_{1} + a_{2} + a_{3} + \cdot \cdot \cdot + a_{300}$ is divisible by 1024
B $a_{0} + a_{2} + a_{4} + \cdot \cdot \cdot + a_{300} = a_{1} + a_{3} + \cdot \cdot \cdot + a_{299}$
D $a_{1} = 100$
$(1 + x + x^{2} + x^{3})^{100}$
$= a_{0} + a_{1} x + a_{2} x^{2} . . . . a_{300} x^{300}$
Now by substituting $x = - 1$ we get
$0 = a_{0} - a_{1} + a_{2} - a_{3} . . . . a_{300}$
Hence
$[a_{0} + a_{2} + a_{4} + . . . a_{2 k} + . . . a_{300}] - [a_{1} + a - 3 + a_{5} + . . . a_{2 k - 1} + . . . a_{299}] = 0$
Hence
$a_{0} + a_{2} + a_{4} + . . . a_{2 k} + . . . a_{300} = a_{1} + a - 3 + a_{5} + . . . a_{2 k - 1} + . . . a_{299}$ ...(i)
Now the term 'x' can be obtained in only one way.
That is
$(x)^{1} (x^{2})^{0} (x^{3})^{0} 1^{99}$
Hence
$a_{1} = \frac{100!}{99! .1! .0! .0!}$
$= 100$
Also
$\sum a_{2 k} = \sum a_{2 k - 1}$ ...from (i)
$= \frac{1}{2} 4^{100}$
$= 2^{200 - 1}$
$= 2^{199}$
Hence it is divisible by $1024 = 2^{10}$ .

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Similar questions

Q. Assertion :Let

(1 + x^{2}) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{36} x^{36}

a_{0} + a_{3} + a_{6} + . . . + a_{36} = \frac{2}{3} (2^{35} + 1)

a_{0} + a_{1} + a_{2} + . . . + a_{36} = 2^{36}

a_{0} + a_{2} + a_{4} + . . . + a_{36} = 2^{35}

(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{10} x^{10}

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} +

(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

$(1 + x + 2 x^{2})^{20}$ = $a_{0} + a_{1} x + a_{2} x^{2}$ ................... $a_{40} x^{40}$ , If $a_{0} + a_{1} + a_{2} + a_{3} . . . . . . . . . . . . . . . . . . a_{40} = 2^{a}$

$a_{0} - a_{1} + a_{2} - a_{3} . . . . . . . . . . . . . . . . . . . . . . a_{40} = 2^{b}$ . Find the value of a + b

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},

find the values of
(i)

a_{0} + a_{1} + a_{2} + \dots . + a_{2 n}

a_{0} - a_{1} + a_{2} - a_{3} \dots . . + a_{2 n}

(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}

, then value of

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

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