Engineering Mathematics

Question

How the value of a1 = 3,  a2 = 2   is calculated.

Consider a sequence $\left\{a_{k}\right\}$ whose generating function is $G(x)=x^{2}+3 x+7+\left(\frac{1}{1-x^{2}}\right) . a_{0}$ and $a_{7}$ are $1^{\text {st }}$ and 8 th term of given sequence respectively. The value of $a_{0}+$ $a_{7}=$ $\qquad$ .

8
Correct Option

Solution :
8
Let,
\[
\begin{aligned}
G(x) & =x^{2}+3 x+7+f(x) \\
f(x) & =\frac{1}{1-x^{2}}=\frac{1}{2}\left[\frac{1}{1-x}+\frac{1}{1+x}\right] \\
& =\frac{1}{2}\left[1+x+x^{2}+x^{3}+\ldots \ldots . .+1-x+x^{2}-x^{3}+\ldots \ldots . .\right] \\
& =\frac{1}{1-x}\left[2+2 x^{2}+2 x^{4}+2 x^{6} \ldots \ldots . .\right] \\
& =1+x^{2}+x^{4}+x^{6}+\ldots \ldots . . \\
G(x) & =7+3 x+x^{2}+f(x) \\
& =7+3 x+x^{2}+1+x^{2}+x^{4}+\ldots \ldots . . \\
& =8 x^{0}+3 x+2 x^{2}+x^{4}+x^{6}+\ldots \ldots . .
\end{aligned}
\]

From generating function is $G(x)$,
\[
\begin{array}{l}
a_{0}=8, a_{1}=3, a_{2}=2, a_{3}=0, a_{4}=1, a_{5}=0, a_{6}=1 \\
a_{n}=\left\{\begin{array}{l}
0, \text { if } n>2 \text { and } n \text { is odd } \\
1, \text { if } n>2 \text { and } n \text { is even }
\end{array}\right.
\end{array}
\]

So,
\[
\begin{aligned}
a_{7} & =0 \\
a_{0}+a_{7} & =8+0=8
\end{aligned}
\]

Comments

We know that ;

$\frac{1}{1-x}=1+ x+x^{2}+x^{3}+x^{4}+x^{5}+...................$

Putting $x^{2}$ in place of x we get ,

$\frac{1}{1-x^{2}}=1+ x^{2}+x^{4}+x^{6}+x^{8}+x^{10}+...................$

Given ,

$G(x)=x^{2}+3x+7+\frac{1}{1-x^{2}}$

 $G(x)=x^{2}+3x+7+1+x^2+x^4+x^{6}+x^8$

 $G(x)=8+3x+2x^2+0.x^3+x^4+0.x^5+x^{6}+0.x^7+x^8+x^{10}+x^{12}+........$

So, $a_{0}=8$ , $a_{7}=0$    ;  $a_{0}$, $a_{7}$ are coefficient of $x^0$ and $x^7$ .

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thank you @Kabir5454 I got the solution. 👌

 

 

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@GateOverflow04  Atleast someone noticed😂😂😂