
If $$\left( 1+x-2x^{2}\right)^{6} =1+a_{1}x+a_{2}x^{2}+\cdots +a_{12}x^{12}$$, then the expansion $$a_{2}+a_{4}+a_{6}+\cdots +a_{12}$$ has the value
A) 32
B) 63
C) 64
D) 31
Answer
608.1k+ views
Hint: In this question it is given that If $$\left( 1+x-2x^{2}\right)^{6} =1+a_{1}x+a_{2}x^{2}+\cdots +a_{12}x^{12}$$.......(1)
Then we have to find the value of $$a_{2}+a_{4}+a_{6}+\cdots +a_{12}$$ . So to find the solution we have to put the value of x=1 and then x=-1 in the equation (1), then this will give us two more equations, from which we are able to find our required solution.
Complete step-by-step solution:
Given equation,
$$\left( 1+x-2x^{2}\right)^{6} =1+a_{1}x+a_{2}x^{2}+\cdots +a_{12}x^{12}$$..............(1)
Now, by putting x=1 in (1) , we get,
$$\left( 1+1-2\times 1^{2}\right)^{6} =1+a_{1}.1+a_{2}.1^{2}+\cdots +a_{12}.1^{12}$$
$$\Rightarrow 0=1+a_{1}+a_{2}+\cdots +a_{12}$$
$$\Rightarrow 1+a_{1}+a_{2}+\cdots +a_{12}=0$$............(2)
Now we are going to put x=-1 in (1),
$$ \left\{ 1+\left( -1\right) -2.\left( -1\right)^{2} \right\}^{6} =1+a_{1}.\left( -1\right) +a_{2}.\left( -1\right)^{2} +\cdots +a_{12}.\left( -1\right)^{12}$$
$$\Rightarrow \left\{ 1-1-2\right\}^{6} =1-a_{1}+a_{2}-a_{3}+\cdots -a_{11}+a_{12}$$
$$\Rightarrow 2^{6}=1-a_{1}+a_{2}-a_{3}+\cdots -a_{11}+a_{12}$$
$$\Rightarrow 1-a_{1}+a_{2}-a_{3}+\cdots -a_{11}+a_{12}=2^{6}$$.......(3)
Now by adding equation (1) and (2) we get,
$$ (1+a_{1}+a_{2}+a_{3}+\cdots +a_{12})+(1-a_{1}+a_{2}-a_{3}+\cdots +a_{12})=0+2^{6}$$
$$\Rightarrow \left( 2+2a_{2}+2a_{4}+2a_{6}+\cdots +2a_{12}\right) =2^{6}$$
$$\Rightarrow 2\left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =2^{6}$$
$$\Rightarrow \left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =\dfrac{2^{6}}{2}$$
$$\Rightarrow \left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =2^{\left( 6-1\right) }$$ [$$\because \dfrac{a^{n}}{a^{m}} =a^{n-m}$$]
$$\Rightarrow \left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =2^{5}$$
$$\Rightarrow a_{2}+a_{4}+a_{6}\cdots +a_{12}=2^{5}-1$$
$$\Rightarrow a_{2}+a_{4}+a_{6}\cdots +a_{12}=32-1$$
$$\Rightarrow a_{2}+a_{4}+a_{6}\cdots +a_{12}=31$$
Hence the correct option is option D.
Note: To solve this type of question you have to put different values of x in equation (1), since in this question it only asked to find the result of the expansion $$a_{2}+a_{4}+a_{6}+\cdots +a_{12}$$, where the constant coefficient of every term is 1, so for this the only possible values of x which we can put is 1 or -1.
Then we have to find the value of $$a_{2}+a_{4}+a_{6}+\cdots +a_{12}$$ . So to find the solution we have to put the value of x=1 and then x=-1 in the equation (1), then this will give us two more equations, from which we are able to find our required solution.
Complete step-by-step solution:
Given equation,
$$\left( 1+x-2x^{2}\right)^{6} =1+a_{1}x+a_{2}x^{2}+\cdots +a_{12}x^{12}$$..............(1)
Now, by putting x=1 in (1) , we get,
$$\left( 1+1-2\times 1^{2}\right)^{6} =1+a_{1}.1+a_{2}.1^{2}+\cdots +a_{12}.1^{12}$$
$$\Rightarrow 0=1+a_{1}+a_{2}+\cdots +a_{12}$$
$$\Rightarrow 1+a_{1}+a_{2}+\cdots +a_{12}=0$$............(2)
Now we are going to put x=-1 in (1),
$$ \left\{ 1+\left( -1\right) -2.\left( -1\right)^{2} \right\}^{6} =1+a_{1}.\left( -1\right) +a_{2}.\left( -1\right)^{2} +\cdots +a_{12}.\left( -1\right)^{12}$$
$$\Rightarrow \left\{ 1-1-2\right\}^{6} =1-a_{1}+a_{2}-a_{3}+\cdots -a_{11}+a_{12}$$
$$\Rightarrow 2^{6}=1-a_{1}+a_{2}-a_{3}+\cdots -a_{11}+a_{12}$$
$$\Rightarrow 1-a_{1}+a_{2}-a_{3}+\cdots -a_{11}+a_{12}=2^{6}$$.......(3)
Now by adding equation (1) and (2) we get,
$$ (1+a_{1}+a_{2}+a_{3}+\cdots +a_{12})+(1-a_{1}+a_{2}-a_{3}+\cdots +a_{12})=0+2^{6}$$
$$\Rightarrow \left( 2+2a_{2}+2a_{4}+2a_{6}+\cdots +2a_{12}\right) =2^{6}$$
$$\Rightarrow 2\left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =2^{6}$$
$$\Rightarrow \left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =\dfrac{2^{6}}{2}$$
$$\Rightarrow \left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =2^{\left( 6-1\right) }$$ [$$\because \dfrac{a^{n}}{a^{m}} =a^{n-m}$$]
$$\Rightarrow \left( 1+a_{2}+a_{4}+a_{6}+\cdots +a_{12}\right) =2^{5}$$
$$\Rightarrow a_{2}+a_{4}+a_{6}\cdots +a_{12}=2^{5}-1$$
$$\Rightarrow a_{2}+a_{4}+a_{6}\cdots +a_{12}=32-1$$
$$\Rightarrow a_{2}+a_{4}+a_{6}\cdots +a_{12}=31$$
Hence the correct option is option D.
Note: To solve this type of question you have to put different values of x in equation (1), since in this question it only asked to find the result of the expansion $$a_{2}+a_{4}+a_{6}+\cdots +a_{12}$$, where the constant coefficient of every term is 1, so for this the only possible values of x which we can put is 1 or -1.
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