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7.Binomial Theorem
hard
${\left( {1 + {x^n} + {x^{253}}} \right)^{10}}$ ના વિસ્તરણમાં $x^{1012}$ સહગુણક કેટલો થાય ? (જ્યાં $n \leq 22$ એ કોઈ પણ ધન પૃણાંક છે )
A
$1$
B
$^{10}{C_4}$
C
$4n$
D
$^{253}{C_4}$
(JEE MAIN-2014)
Solution
Given expansion $\left(1+x^{n}+x^{253}\right)^{10}$
Let $x^{1012}=(1)^{a}\left(x^{n}\right)^{b} \cdot\left(x^{253}\right)^{c}$
Here $a, b, c, n$ are all $+ve$ integers and
$a$ $\leq 10, b \leq 10, c \leq 4, n \leq 22, a+b+c=10$
Now $b n+253 c=1012$
$\Rightarrow b n=253(4-c)$
For $c<4$ and $n \leq 22 ; b>10,$ which is not possible.
$\therefore c=4, b=0, a=6$
$\therefore x^{1012}=(1)^{6} \cdot\left(x^{12}\right)^{0} \cdot\left(x^{253}\right)^{4}$
Hence the coefficient of $x^{1012}$
$=\frac{10 !}{6 ! 0 ! 4 !}=10 \mathrm{C}_{4}$
Standard 11
Mathematics