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

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Solution

When we put x = 1, we get the first series and when we put x = -1, we get second series

⇒ $(1 + 1 + 2)^{20}$ = $a_{0} + a_{1} + a_{2} . . . . . . . . a_{40}$

= $4^{20}$ = $2^{40}$

= $2^{a}$

⇒ a = 40

Also $(1 - 1 + 2)^{20}$ = $a_{0} - a_{1} + a_{2} . . . . . . . . . . . . . . . . . a_{40}$ = $2^{20}$

⇒ b = 20 ⇒ a + b = 60

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Similar questions

{(1 - x + x^{4})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{40} x^{40}

{(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{39} x^{39} + a_{40} x^{40}

a_{0} + a_{1} + a_{2} + . . . . . . + a_{38}

(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . a_{40} x^{40}

a_{1} + a_{3} + a_{5} + . . . . a_{39}

{(1 + x - 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}

a_{1} + a_{3} + a_{5} + \dots + a_{39} = - 2^{k}

k =

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

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

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

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