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7.Binomial Theorem
hard
यदि धन पूर्णाकों $m$ तथा $n$ के लिए
$(1-y)^{m}(1+y)^{n}=1+a_{1} y+a_{2} y^{2}+\ldots .+a_{m-n} y^{m+n}$ तथा $a_{1}=a_{2}=10$ हैं, तो $(m+n)$ बराबर है
A
$88$
B
$64$
C
$100$
D
$80$
(JEE MAIN-2021)
Solution
$(1-y)^{m}(1+y)^{n}$
Coefficient of $y=1 .{ }^{n} C_{1}+{ }^{m} C_{1}(-1)$
$=n-m=10$ $\ldots(1)$
Coefficient of $\mathrm{y}^{2}\left(\mathrm{a}_{2}\right)$
$=1 .{ }^{n} \mathrm{C}_{2}-{ }^{n} \mathrm{C}_{1} \cdot{ }^{n} C_{1 .}+1 \cdot{ }^{m} C_{2}=10$
$=\frac{n(n-1)}{2}-m \cdot n+\frac{m(m-1)}{2}=10$
$m^{2}+n^{2}-2 m n-(n+m)=20$
$n+m=80$
$(n-m)^{2}-(n+m)=20$
By equation $(1)\, \,(2)$
$\mathrm{m}=35, n=45$
Standard 11
Mathematics
