Here we can find the coefficient of
x53 in the given expression
We know that
Binomial theorem for positive integral index
If a , b are any two real numbers and n any natural number then
(a+b)n =
C0anb° +
C1an−1b +
C2an−2b2 + ... +
Cran−rbr + ... +
Cnx0bn ,
where
Cr =
Here
C0,C1,C2 ...
Cn are called binomial coefficients
i.e.,
(a+b)n =
Cran−rbr Properties of binomial coefficients
In the binomial expansion of (1 + x) n the coefficients
C0,C1,C2 , ... ,
Cn are denoted by
C0,C1,C2,...,Cn respectively
C0+C1+C2+...+Cn =
2n We have
(1+x)50 (1+x)50 =
Crxr (Since by binomial expansion)
∴ sum of the coefficients of odd power of x
=
C1+C3 + ... +
C49 =
(Since
C0+C2+C4 + ... =
C1+C3+C5 + ... = 2n - 1)
=
×250 (since
C0+C1+C2 + ... +
C50 =
2n)
=
250×2−1 =
250−1 =
249 Hence,
In the expansion of
(1+x)50 the sum of the cofficients of the odd powers of x is
249 Another method :
We know that
To find the sum of coefficients of odd powers of x, we find the values of the expression by replacing the variables by 1 and - 1 respectively and the subtracting the two results
Let f (x) =
(1+x)50 Replacing the variables by 1 and - 1
f (1) =
(1+1)50 =
250 f (2) =
(1−1)50 = 0
By subtracting the two results we get the odd power of x
=
=
=
250×2−1 =
249 Hence
In the expansion of
(1+x)50 the sum of the coefficients of the odd powers of x is
249.