If 1-x-x2 8 - a0 - a1x - a2x2 -a16x16 for all real x- then a5 is equal to

T_r=8!a! b! c!(1)^a(x)^b(x^2)^c, where a+b+c=8 For coefficient of x^5, nbsp;b+2c=5 (b,c)=(1,2),(3,1),(5,0) ∴ a_5=8!5!·1!·2!+8!4!·3!·1!+8!3!·5!·0! ⇒ a_5=168+2 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Binomial Coefficients with Alternate Signs

If 1+x+x2 8 =...

Question

If ${(1 + x + x^{2})}^{8} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{16} x^{16}$ for all real $x,$ then $a_{5}$ is equal to

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Solution

$T_{r} = \frac{8!}{a! b! c!} (1)^{a} (x)^{b} (x^{2})^{c},$ where $a + b + c = 8$
For coefficient of $x^{5},$ $b + 2 c = 5$
$(b, c) = (1, 2), (3, 1), (5, 0)$

$∴ a_{5} = \frac{8!}{5! \cdot 1! \cdot 2!} + \frac{8!}{4! \cdot 3! \cdot 1!} + \frac{8!}{3! \cdot 5! \cdot 0!}$
$\Rightarrow a_{5} = 168 + 280 + 56 = 504$

Alternate Solution :
$[(1 + x) + x^{2}]^{8} =^{8} C_{0} (1 + x)^{8} +^{8} C_{1} (x^{2})^{1} (1 + x)^{7} +^{8} C_{2} (x^{2})^{2} (1 + x)^{6} +^{8} C_{3} (x^{2})^{3} (1 + x)^{5} + \dots$
First three terms consist $x^{5} .$
So, $a_{5} =^{8} C_{0} \cdot^{8} C_{5} +^{8} C_{1} \cdot^{7} C_{3} +^{8} C_{2} \cdot^{6} C_{1}$
$\Rightarrow a_{5} = 56 + 280 + 168 = 504$

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If (1+x+x2)8=a0+a1x+a2x2+⋯+a16x16 for all real x, then a5 is equal to

If ${(1 + x + x^{2})}^{8} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{16} x^{16}$ for all real $x,$ then $a_{5}$ is equal to