Answer :
To solve for the term independent of [tex]\( x \)[/tex] in the expansion of [tex]\(\left(2x - \frac{5}{x}\right)^6\)[/tex], we'll utilize the Binomial Theorem. The Binomial Theorem states:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
For our specific polynomial [tex]\(\left(2x - \frac{5}{x}\right)^6\)[/tex], we can set [tex]\( a = 2x \)[/tex] and [tex]\( b = -\frac{5}{x} \)[/tex], with [tex]\( n = 6 \)[/tex].
We need to identify the term in the expansion where the power of [tex]\( x \)[/tex] is zero. The general term in this expansion is:
[tex]\[ T_k = \binom{6}{k} (2x)^{6-k} \left(-\frac{5}{x}\right)^k \][/tex]
Simplifying the powers of [tex]\( x \)[/tex]:
[tex]\[ T_k = \binom{6}{k} 2^{6-k} x^{6-k} \left(-\frac{5}{x}\right)^k = \binom{6}{k} 2^{6-k} (-5)^k x^{6-k-k} = \binom{6}{k} 2^{6-k} (-5)^k x^{6-2k} \][/tex]
We want the exponent of [tex]\( x \)[/tex] to be zero:
[tex]\[ 6 - 2k = 0 \implies 2k = 6 \implies k = 3 \][/tex]
So, we are interested in the term where [tex]\( k = 3 \)[/tex]. Substitute [tex]\( k = 3 \)[/tex] back into the general term expression:
[tex]\[ T_3 = \binom{6}{3} 2^{6-3} (-5)^3 x^0 = \binom{6}{3} 2^3 (-5)^3 \][/tex]
Calculate each part individually:
- Binomial coefficient [tex]\(\binom{6}{3} = 20\)[/tex]
- [tex]\(2^3 = 8\)[/tex]
- [tex]\((-5)^3 = -125\)[/tex]
Now, multiply these results together:
[tex]\[ T_3 = 20 \cdot 8 \cdot (-125) = 20 \cdot (-1000) = -20000 \][/tex]
Therefore, the term independent of [tex]\( x \)[/tex] in the expansion of [tex]\(\left(2x - \frac{5}{x}\right)^6\)[/tex] is [tex]\(-20000\)[/tex]. Thus, the correct answer is:
[tex]\[ \boxed{-20000} \][/tex]
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
For our specific polynomial [tex]\(\left(2x - \frac{5}{x}\right)^6\)[/tex], we can set [tex]\( a = 2x \)[/tex] and [tex]\( b = -\frac{5}{x} \)[/tex], with [tex]\( n = 6 \)[/tex].
We need to identify the term in the expansion where the power of [tex]\( x \)[/tex] is zero. The general term in this expansion is:
[tex]\[ T_k = \binom{6}{k} (2x)^{6-k} \left(-\frac{5}{x}\right)^k \][/tex]
Simplifying the powers of [tex]\( x \)[/tex]:
[tex]\[ T_k = \binom{6}{k} 2^{6-k} x^{6-k} \left(-\frac{5}{x}\right)^k = \binom{6}{k} 2^{6-k} (-5)^k x^{6-k-k} = \binom{6}{k} 2^{6-k} (-5)^k x^{6-2k} \][/tex]
We want the exponent of [tex]\( x \)[/tex] to be zero:
[tex]\[ 6 - 2k = 0 \implies 2k = 6 \implies k = 3 \][/tex]
So, we are interested in the term where [tex]\( k = 3 \)[/tex]. Substitute [tex]\( k = 3 \)[/tex] back into the general term expression:
[tex]\[ T_3 = \binom{6}{3} 2^{6-3} (-5)^3 x^0 = \binom{6}{3} 2^3 (-5)^3 \][/tex]
Calculate each part individually:
- Binomial coefficient [tex]\(\binom{6}{3} = 20\)[/tex]
- [tex]\(2^3 = 8\)[/tex]
- [tex]\((-5)^3 = -125\)[/tex]
Now, multiply these results together:
[tex]\[ T_3 = 20 \cdot 8 \cdot (-125) = 20 \cdot (-1000) = -20000 \][/tex]
Therefore, the term independent of [tex]\( x \)[/tex] in the expansion of [tex]\(\left(2x - \frac{5}{x}\right)^6\)[/tex] is [tex]\(-20000\)[/tex]. Thus, the correct answer is:
[tex]\[ \boxed{-20000} \][/tex]
