Question

If $\sf{{(x+1)}^{7} = {a}_{7}{x}^{7} + {a}_{6}{x}^{6} + {a}_{5}{x}^{5} + ... + {a}_{1}x + {a}_{0}}$
$\sf{and \:{x}_{1}, {x}_{2}, ... , {x}_{7}}$ are the zeroes of $\sf{{(x+1)}^{7}}$.
$\bf{\underline{Find the value of :-}}$
(i) $\sf{\sum{x}_{1} {x}_{2}}$
(ii) $\sf{{a}_{6} + {a}_{4} + {a}_{2} + {a}_{0}}$

[Ans. (i) = 21 and (ii) = 64]

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Answer 1

We're given,

$\sf{\longrightarrow (x+1)^7=a_7x^7+a_6x^6+a_5x^5+\,\dots\,+a_1x+a_0}$

Or,

$\displaystyle\sf{\longrightarrow (x+1)^7=\sum_{r=0}^7a_{7-r}x^{7-r}}$

By binomial expansion,

$\displaystyle\sf{\longrightarrow \sum_{r=0}^7\,^7C_r\,x^{7-r}=\sum_{r=0}^7a_{7-r}x^{7-r}}$

From this we get,

$\sf{\longrightarrow a_{7-r}=\,^7C_r\quad\forall r\in\{0,\ 1,\ 2,\,\dots\,\ 7\}\quad\quad\dots(1)}$

(i)  The zero of $\sf{(x+1)^7=0}$ is only $\sf{x=-1.}$ Thus,

$\sf{\longrightarrow x_1=x_2=x_3=\,\dots\,=x_7=-1}$

The sum $\displaystyle\sf{\sum x_1x_2}$ actually is the sum of all possible combinations of the zeroes $\sf{x_1,\ x_2,\ x_3,\,\dots\,,\ x_7,}$ two taken at a time.

The no. of possible combinations of taking two among the 7 zeroes is $\sf{^7C_2.}$

Therefore,

$\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{(-1)(-1)+(-1)(-1)+(-1)(-1)+\,\dots\,+(-1)(-1)}_{\sf{\,^7C_2\ terms}}}$

$\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{1+1+1+\,\dots\,+1}_{\sf{\,^7C_2\ terms}}}$

$\displaystyle\sf{\longrightarrow \sum x_1x_2=\,^7C_2}$

$\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}$

OR

The expansion of $\sf{(x+1)^7}$ implies,

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_5}{a_7}}$

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_{7-2}}{a_{7-0}}}$

From (1),

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{^7C_2}{^7C_0}}$

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{21}{1}}$

$\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}$

(ii)  Put $\sf{x=1}$ in the expansion.

$\sf{\longrightarrow (1+1)^7=a_7(1)^7+a_6(1)^6+a_5(1)^5+\,\dots\,+a_1(1)+a_0}$

$\sf{\longrightarrow2^7=a_7+a_6+a_5+\,\dots\,+a_1+a_0\quad\quad\dots(2)}$

Similarly, put $\sf{x=-1.}$

$\sf{\longrightarrow (-1+1)^7=a_7(-1)^7+a_6(-1)^6+a_5(-1)^5+\,\dots\,+a_1(-1)+a_0}$

$\sf{\longrightarrow 0=-a_7+a_6-a_5+\,\dots\,-a_1+a_0\quad\quad\dots(3)}$

Adding (2) and (3),

$\sf{\longrightarrow2^7+0=(a_7+a_6+a_5+\,\dots\,+a_1+a_0)+(-a_7+a_6-a_5+\,\dots\,-a_1+a_0)}$

$\sf{\longrightarrow2^7=a_7+a_6+a_5+\,\dots\,+a_1+a_0-a_7+a_6-a_5+\,\dots\,-a_1+a_0}$

$\sf{\longrightarrow2^7=2a_6+2a_4+2a_2+2a_0}$

$\sf{\longrightarrow2^7=2(a_6+a_4+a_2+a_0)}$

$\sf{\longrightarrow a_6+a_4+a_2+a_0=\dfrac{2^7}{2}}$

$\sf{\longrightarrow a_6+a_4+a_2+a_0=2^6}$

$\sf{\longrightarrow\underline{\underline{a_6+a_4+a_2+a_0=64}}}$

Answer 2

We're given,

$\sf{\longrightarrow (x+1)^7=a_7x^7+a_6x^6+a_5x^5+\,\dots\,+a_1x+a_0}$

Or,

$\displaystyle\sf{\longrightarrow (x+1)^7=\sum_{r=0}^7a_{7-r}x^{7-r}}$

By binomial expansion,

$\displaystyle\sf{\longrightarrow \sum_{r=0}^7\,^7C_r\,x^{7-r}=\sum_{r=0}^7a_{7-r}x^{7-r}}$

From this we get,

$\sf{\longrightarrow a_{7-r}=\,^7C_r\quad\forall r\in\{0,\ 1,\ 2,\,\dots\,\ 7\}\quad\quad\dots(1)}$

(i)  The zero of $\sf{(x+1)^7=0}$ is only $\sf{x=-1.}$ Thus,

$\sf{\longrightarrow x_1=x_2=x_3=\,\dots\,=x_7=-1}$

The sum $\displaystyle\sf{\sum x_1x_2}$ actually is the sum of all possible combinations of the zeroes $\sf{x_1,\ x_2,\ x_3,\,\dots\,,\ x_7,}$ two taken at a time.

The no. of possible combinations of taking two among the 7 zeroes is $\sf{^7C_2.}$

Therefore,

$\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{(-1)(-1)+(-1)(-1)+(-1)(-1)+\,\dots\,+(-1)(-1)}_{\sf{\,^7C_2\ terms}}}$

$\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{1+1+1+\,\dots\,+1}_{\sf{\,^7C_2\ terms}}}$

$\displaystyle\sf{\longrightarrow \sum x_1x_2=\,^7C_2}$

$\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}$

OR

The expansion of $\sf{(x+1)^7}$ implies,

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_5}{a_7}}$

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_{7-2}}{a_{7-0}}}$

From (1),

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{^7C_2}{^7C_0}}$

$\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{21}{1}}$

$\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}$

(ii)  Put $\sf{x=1}$ in the expansion.

$\sf{\longrightarrow (1+1)^7=a_7(1)^7+a_6(1)^6+a_5(1)^5+\,\dots\,+a_1(1)+a_0}$

$\sf{\longrightarrow2^7=a_7+a_6+a_5+\,\dots\,+a_1+a_0\quad\quad\dots(2)}$

Similarly, put $\sf{x=-1.}$

$\sf{\longrightarrow (-1+1)^7=a_7(-1)^7+a_6(-1)^6+a_5(-1)^5+\,\dots\,+a_1(-1)+a_0}$

$\sf{\longrightarrow 0=-a_7+a_6-a_5+\,\dots\,-a_1+a_0\quad\quad\dots(3)}$

Adding (2) and (3),

$\sf{\longrightarrow2^7+0=(a_7+a_6+a_5+\,\dots\,+a_1+a_0)+(-a_7+a_6-a_5+\,\dots\,-a_1+a_0)}$

$\sf{\longrightarrow2^7=a_7+a_6+a_5+\,\dots\,+a_1+a_0-a_7+a_6-a_5+\,\dots\,-a_1+a_0}$

$\sf{\longrightarrow2^7=2a_6+2a_4+2a_2+2a_0}$

$\sf{\longrightarrow2^7=2(a_6+a_4+a_2+a_0)}$

$\sf{\longrightarrow a_6+a_4+a_2+a_0=\dfrac{2^7}{2}}$

$\sf{\longrightarrow a_6+a_4+a_2+a_0=2^6}$

$\sf{\longrightarrow\underline{\underline{a_6+a_4+a_2+a_0=64}}}$

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