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  • Let [x] denote greatest integer less than or equal to x. If for <im

Let [x] denote greatest integer less than or equal to x. If for n \in N \text {, }\left(1-x+x^3\right)^n=\sum_{j=0}^{3 n} a_j x^j \text {, then } \sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+5 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1} \text { is equal to }

Option: 1

0


Option: 2

-1


Option: 3

1


Option: 4

2


Answers (1)

We have,

\left(1-x+x^3\right)^n=a_0 x^0+a_1 x+a_2 x^2+\ldots \ldots+a_{3 n} x^{3 n}....(i)

Putting x = 1 in (i), we get...(ii)

a_0+a_1+a_2+\ldots \ldots+a_{3 n}=1...(ii)

Also, putting x = –1 in (i), we get

a_0-a_1+a_2-a_3+\ldots \ldots+(-1)^{3 n} \cdot a_{3 n}=1...(iii)

Now, adding and subtracting (ii) and (iii), we get

\begin{aligned} & 2\left\{a_0+a_2+a_4+\ldots .+a_2\left[\frac{3 n}{2}\right]\right\}=2 \\ & \Rightarrow \sum_{j=0}^{[3 n / 2]} a_{2 j}=1 \\ & \text { and } 2\left[a_1+a_3+a_5+\ldots . .+a_{\left.2\left[\frac{3 n-1}{2}\right]+1\right\}}\right\}=0 \\ & \Rightarrow\left[\frac{3 n-1}{\left.\sum_{j=0}^2\right]} a_{2 j+1}=0\right. \end{aligned}....(v)

\therefore \quad \sum_{j=0}^{[3 n / 2]} a_{2 j}+5 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}=1+5 \times 0=1(Using (iv) and (v))

Posted by

Sumit Saini

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