Question

Coefficient of $x^{29}$ in the polynomial $(x-\frac{1}{1\times 3})(x-\frac{2}{1\times 3\times 5})(x-\frac{3}{1\times 3\times 5\times 7})............(x-\frac{30}{1\times 3\times ......\times 61})$ is

• A. $\frac{1}{2}(1-\frac{1}{1\times 3\times ....\times 61})$
• B. $-\frac{1}{2}+\frac{1}{1\times 3\times ...\times 61}$
• C. $-\frac{1}{2}(1-\frac{1}{1\times 3\times ....\times 61})$
• D. $\frac{29}{1\times 3\times ......\times 61}$

Answer 1

The correct option is C $-\frac{1}{2}(1-\frac{1}{1\times 3\times ....\times 61})$

Given that

$(x-\frac{1}{1-3})(x-\frac{2}{1.25})(x-\frac{3}{1.357}).......(x-\frac{30}{1.361})$

Then the polynomial

$(x-a)(x-b)(x-c)------(x-z)$

is explained as

$x^{\alpha }-a_1{x}^{\alpha -1}+a_2{x}^{\alpha -2}------a_{31}{x}^{\alpha -\alpha }$

when

$\alpha =$total numbers of brackets

$a_1=a+b+b+----+z$

$a_2=(ab+ac+ad-------az)+(bc+bd+bc+------+(yz)$

now, $a_1=a,b,c,d------------2$

we have to find coefficient of $x^{29}$

Hence $x^{29}3$ the $2^{nd}$ term whose coefficient is $a_1$.

$\therefore a_1=(\frac{1}{1.3}+\frac{2}{\mathrm{1.3.5}}+----\frac{r}{\mathrm{1.3.5}----\left(2r\right)})$

$\therefore a_1={\sum }_{r=1}^{30}\frac{r}{\mathrm{1.3.5}---(2r+1)}$

$\therefore a_1=\frac{1}{2}\sum _{r=1}^{30}\frac{2r+1-1}{\mathrm{1.3.5}---(2r+1)}$

$\therefore a_1=\frac{1}{2}\sum _{r=1}^{30}[\left(\frac{1}{1.3---(2r-1)}\right)-\left(\frac{1}{1.3---(2r+1)}\right)]$

If is of the form

$\sum _{r=1}^{30}(t_r-t_{rn})$

Hence,

$a_1=\frac{1}{2}(t_1-t_{31})$

Hence, the final answer is

$-a_1=\frac{-1}{2}(1-\frac{1}{\mathrm{1.3.5.}---61})$

This is the correct answer.