If [tex]\sf{{(x+1)}^{7} = {a}_{7}{x}^{7} + {a}_{6}{x}^{6} + {a}_{5}{x}^{5} + ... + {a}_{1}x + {a}_{0}}[/tex]
[tex]\sf{and \:{x}_{1}, {x}_{2}, ... , {x}_{7}}[/tex] are the zeroes of [tex]\sf{{(x+1)}^{7}}[/tex].
[tex]\bf{\underline{Find the value of :-}}[/tex]
(i) [tex]\sf{\sum{x}_{1} {x}_{2}}[/tex]
(ii) [tex]\sf{{a}_{6} + {a}_{4} + {a}_{2} + {a}_{0}}[/tex]
[Ans. (i) = 21 and (ii) = 64]
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We're given,
[tex]\sf{\longrightarrow (x+1)^7=a_7x^7+a_6x^6+a_5x^5+\,\dots\,+a_1x+a_0}[/tex]
Or,
[tex]\displaystyle\sf{\longrightarrow (x+1)^7=\sum_{r=0}^7a_{7-r}x^{7-r}}[/tex]
By binomial expansion,
[tex]\displaystyle\sf{\longrightarrow \sum_{r=0}^7\,^7C_r\,x^{7-r}=\sum_{r=0}^7a_{7-r}x^{7-r}}[/tex]
From this we get,
[tex]\sf{\longrightarrow a_{7-r}=\,^7C_r\quad\forall r\in\{0,\ 1,\ 2,\,\dots\,\ 7\}\quad\quad\dots(1)}[/tex]
(i) The zero of [tex]\sf{(x+1)^7=0}[/tex] is only [tex]\sf{x=-1.}[/tex] Thus,
[tex]\sf{\longrightarrow x_1=x_2=x_3=\,\dots\,=x_7=-1}[/tex]
The sum [tex]\displaystyle\sf{\sum x_1x_2}[/tex] actually is the sum of all possible combinations of the zeroes [tex]\sf{x_1,\ x_2,\ x_3,\,\dots\,,\ x_7,}[/tex] two taken at a time.
The no. of possible combinations of taking two among the 7 zeroes is [tex]\sf{^7C_2.}[/tex]
Therefore,
[tex]\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{(-1)(-1)+(-1)(-1)+(-1)(-1)+\,\dots\,+(-1)(-1)}_{\sf{\,^7C_2\ terms}}}[/tex]
[tex]\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{1+1+1+\,\dots\,+1}_{\sf{\,^7C_2\ terms}}}[/tex]
[tex]\displaystyle\sf{\longrightarrow \sum x_1x_2=\,^7C_2}[/tex]
[tex]\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}[/tex]
OR
The expansion of [tex]\sf{(x+1)^7}[/tex] implies,
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_5}{a_7}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_{7-2}}{a_{7-0}}}[/tex]
From (1),
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{^7C_2}{^7C_0}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{21}{1}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}[/tex]
(ii) Put [tex]\sf{x=1}[/tex] in the expansion.
[tex]\sf{\longrightarrow (1+1)^7=a_7(1)^7+a_6(1)^6+a_5(1)^5+\,\dots\,+a_1(1)+a_0}[/tex]
[tex]\sf{\longrightarrow2^7=a_7+a_6+a_5+\,\dots\,+a_1+a_0\quad\quad\dots(2)}[/tex]
Similarly, put [tex]\sf{x=-1.}[/tex]
[tex]\sf{\longrightarrow (-1+1)^7=a_7(-1)^7+a_6(-1)^6+a_5(-1)^5+\,\dots\,+a_1(-1)+a_0}[/tex]
[tex]\sf{\longrightarrow 0=-a_7+a_6-a_5+\,\dots\,-a_1+a_0\quad\quad\dots(3)}[/tex]
Adding (2) and (3),
[tex]\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)}[/tex]
[tex]\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}[/tex]
[tex]\sf{\longrightarrow2^7=2a_6+2a_4+2a_2+2a_0}[/tex]
[tex]\sf{\longrightarrow2^7=2(a_6+a_4+a_2+a_0)}[/tex]
[tex]\sf{\longrightarrow a_6+a_4+a_2+a_0=\dfrac{2^7}{2}}[/tex]
[tex]\sf{\longrightarrow a_6+a_4+a_2+a_0=2^6}[/tex]
[tex]\sf{\longrightarrow\underline{\underline{a_6+a_4+a_2+a_0=64}}}[/tex]
We're given,
[tex]\sf{\longrightarrow (x+1)^7=a_7x^7+a_6x^6+a_5x^5+\,\dots\,+a_1x+a_0}[/tex]
Or,
[tex]\displaystyle\sf{\longrightarrow (x+1)^7=\sum_{r=0}^7a_{7-r}x^{7-r}}[/tex]
By binomial expansion,
[tex]\displaystyle\sf{\longrightarrow \sum_{r=0}^7\,^7C_r\,x^{7-r}=\sum_{r=0}^7a_{7-r}x^{7-r}}[/tex]
From this we get,
[tex]\sf{\longrightarrow a_{7-r}=\,^7C_r\quad\forall r\in\{0,\ 1,\ 2,\,\dots\,\ 7\}\quad\quad\dots(1)}[/tex]
(i) The zero of [tex]\sf{(x+1)^7=0}[/tex] is only [tex]\sf{x=-1.}[/tex] Thus,
[tex]\sf{\longrightarrow x_1=x_2=x_3=\,\dots\,=x_7=-1}[/tex]
The sum [tex]\displaystyle\sf{\sum x_1x_2}[/tex] actually is the sum of all possible combinations of the zeroes [tex]\sf{x_1,\ x_2,\ x_3,\,\dots\,,\ x_7,}[/tex] two taken at a time.
The no. of possible combinations of taking two among the 7 zeroes is [tex]\sf{^7C_2.}[/tex]
Therefore,
[tex]\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{(-1)(-1)+(-1)(-1)+(-1)(-1)+\,\dots\,+(-1)(-1)}_{\sf{\,^7C_2\ terms}}}[/tex]
[tex]\displaystyle\sf{\longrightarrow \sum x_1x_2=\underbrace{1+1+1+\,\dots\,+1}_{\sf{\,^7C_2\ terms}}}[/tex]
[tex]\displaystyle\sf{\longrightarrow \sum x_1x_2=\,^7C_2}[/tex]
[tex]\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}[/tex]
OR
The expansion of [tex]\sf{(x+1)^7}[/tex] implies,
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_5}{a_7}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{a_{7-2}}{a_{7-0}}}[/tex]
From (1),
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{^7C_2}{^7C_0}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\sum x_1x_2=\dfrac{21}{1}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\underline{\underline{\sum x_1x_2=21}}}[/tex]
(ii) Put [tex]\sf{x=1}[/tex] in the expansion.
[tex]\sf{\longrightarrow (1+1)^7=a_7(1)^7+a_6(1)^6+a_5(1)^5+\,\dots\,+a_1(1)+a_0}[/tex]
[tex]\sf{\longrightarrow2^7=a_7+a_6+a_5+\,\dots\,+a_1+a_0\quad\quad\dots(2)}[/tex]
Similarly, put [tex]\sf{x=-1.}[/tex]
[tex]\sf{\longrightarrow (-1+1)^7=a_7(-1)^7+a_6(-1)^6+a_5(-1)^5+\,\dots\,+a_1(-1)+a_0}[/tex]
[tex]\sf{\longrightarrow 0=-a_7+a_6-a_5+\,\dots\,-a_1+a_0\quad\quad\dots(3)}[/tex]
Adding (2) and (3),
[tex]\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)}[/tex]
[tex]\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}[/tex]
[tex]\sf{\longrightarrow2^7=2a_6+2a_4+2a_2+2a_0}[/tex]
[tex]\sf{\longrightarrow2^7=2(a_6+a_4+a_2+a_0)}[/tex]
[tex]\sf{\longrightarrow a_6+a_4+a_2+a_0=\dfrac{2^7}{2}}[/tex]
[tex]\sf{\longrightarrow a_6+a_4+a_2+a_0=2^6}[/tex]
[tex]\sf{\longrightarrow\underline{\underline{a_6+a_4+a_2+a_0=64}}}[/tex]
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