2-x-x35-a0-a1 x-a15 x15 - find the value of a0-2 a1-a2-2 a3-2 a15A- 1586B- 2048C- 1024D- 1536

The correct option is D 1536 At x=1 a_0+a_1+.....+a_15=4^5.............(1) At x= 1 a_0 a_1+...... a_15=0.................(2) (1) (2)=2a_1+2a_3.......2a_15=4 ...

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Quantitative Aptitude

Series

2+x+x35=a0+a1...

Question

$(2 + x + x^{3})^{5} = a_{0} + a_{1} x + . . . . . + a_{15} x^{15}$ . find the value of $a_{0} + 2 a_{1} + a_{2} + 2 a_{3} . . . . + 2 a_{15}$

1024

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1536

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1586

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2048

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Solution

The correct option is B
1536

At x=1
$a_{0} + a_{1} + . . . . . + a_{15} = 4^{5} . . . . . . . . . . . . . (1)$

At x= -1
$a_{0} - a_{1} + . . . . . . - a_{15} = 0................. (2)$

(1)-(2) $= 2 a_{1} + 2 a_{3} . . . . . . .2 a_{15} = 4^{5} . . . . . . . . . (3)$

$(\frac{[(1) + (2)]}{2}) = a_{0} + a_{2} . . . = \frac{4^{5}}{2} . . . . . . . . (4)$

(3)+(4) $= a_{0} + 2 a_{1} + a_{2} + 2 a_{3} . . . . + 2 a_{15} = 4^{5} + [\frac{4^{5}}{2}] = 1536.$

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(2+x+x3)5=a0+a1x+.....+a15x15. find the value of a0+2a1+a2+2a3....+2a15

The correct option is B 1536 At x=1 a0+a1+.....+a15=45.............(1) At x= -1 a0−a1+......−a15=0.................(2) (1)-(2)=2a1+2a3.......2a15=45.........(3) ([(1)+(2)]2)=a0+a2...=452........(4) (3)+(4)=a0+2a1+a2+2a3....+2a15=45+[452]=1536.

$(2 + x + x^{3})^{5} = a_{0} + a_{1} x + . . . . . + a_{15} x^{15}$ . find the value of $a_{0} + 2 a_{1} + a_{2} + 2 a_{3} . . . . + 2 a_{15}$