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Question: The coefficient of \[{x^{1012}}\] in the expansion of \[{\left( {1 + {x^n} + {x^{253}}} \right)^{10}...

The coefficient of x1012{x^{1012}} in the expansion of (1+xn+x253)10{\left( {1 + {x^n} + {x^{253}}} \right)^{10}},(where n⩽22n \leqslant 22 is any positive integer),is:
A) 1
B) 10C4^{10}{C_4}
C) 4n
D) 253C4^{253}{C_4}

Explanation

Solution

Here first we will assume the given expression to be P and then write its general term using the following formula:-
If the expression is of the form (a+b)n{\left( {a + b} \right)^n} then its general term is:-
Tr+1=nCr(a)n−r(b)r{T_{r + 1}}{ = ^n}{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r} and then we will compute the value of r by comparing and find the coefficient of x1012{x^{1012}}.

Complete step-by-step answer:
Let us assume
P=(1+xn+x253)10P = {\left( {1 + {x^n} + {x^{253}}} \right)^{10}}
It can be written as:-
P=((1+xn)+x253)10P = {\left( {\left( {1 + {x^n}} \right) + {x^{253}}} \right)^{10}}
Now let

a=1+xn b=x253  a = 1 + {x^n} \\\ b = {x^{253}} \\\

And n=10n = 10
Now we will apply the following formula to find the general term of the given expression:-
Tr+1=nCr(a)n−r(b)r{T_{r + 1}}{ = ^n}{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}
Putting in the respective values we get:-
Tr+1=10Cr(1+xn)10−r(x253)r{T_{r + 1}}{ = ^{10}}{C_r}{\left( {1 + {x^n}} \right)^{10 - r}}{\left( {{x^{253}}} \right)^r}
On simplifying we get:-
Tr+1=10Cr(1+xn)10−r(x253r){T_{r + 1}}{ = ^{10}}{C_r}{\left( {1 + {x^n}} \right)^{10 - r}}\left( {{x^{253r}}} \right)……………………………(1)
Now since we need to find the coefficient of x1012{x^{1012}}
Therefore we will equate the power of x in the above equation with 1012 in order to get the value of r
Hence on equating we get:-
253r=1012253r = 1012
Solving for r we get:-

r=1012253 ⇒r=4  r = \dfrac{{1012}}{{253}} \\\ \Rightarrow r = 4 \\\

Putting this value in equation 1 we get:-
Tr+1=10C4(1+xn)10−r(x253(4)){T_{r + 1}}{ = ^{10}}{C_4}{\left( {1 + {x^n}} \right)^{10 - r}}\left( {{x^{253\left( 4 \right)}}} \right)
Simplifying it we get:-
Tr+1=10C4(1+xn)10−r(x1012){T_{r + 1}}{ = ^{10}}{C_4}{\left( {1 + {x^n}} \right)^{10 - r}}\left( {{x^{1012}}} \right)
Hence the coefficient of x1012{x^{1012}} is 10C4^{10}{C_4}

Therefore option B is the correct option.

Note: Students should note that coefficient is the constant term with which a variable term is multiplied.
Also, students should keep the formula for the general term of a binomial term in mind in order to get the correct answer.
The general binomial expansion of (a+b)n{\left( {a + b} \right)^n} is given by:-
(a+b)n=nC0(a)0(b)n+nC1(a)1(b)n−1+nC3(a)3(b)n−3.................................+nCn(a)n(b)0{\left( {a + b} \right)^n}{ = ^n}{C_0}{\left( a \right)^0}{\left( b \right)^n}{ + ^n}{C_1}{\left( a \right)^1}{\left( b \right)^{n - 1}}{ + ^n}{C_3}{\left( a \right)^3}{\left( b \right)^{n - 3}}.................................{ + ^n}{C_n}{\left( a \right)^n}{\left( b \right)^0}