Solution:
If we have to expand the expression
[tex](a - b)^n=_{0}^{n}\textrm{C} a^n - _{1}^{n}\textrm{C} a^{n-1} b+_{2}^{n}\textrm{C}a^{n-2}b^2-_{3}^{n}\textrm{C}a^{n-3}b^3+...............................\pm _{n}^{n}\textrm{C} b^n[/tex]
Whether the sign before last term is positive or negative, Depends on whether , n is even or odd.
Let, [tex]T_{r+1}[/tex] be the term containing ,[tex]x^5 y^5[/tex] in the binomial expansion of [tex](2x-3y)^{10}[/tex].
[tex]T_{r+1}=_{r}^{10}\textrm{C} (2x)^r\times (-3y)^{10-r}[/tex]
If we replace r, by 5 in the above equation we get coefficient of ,[tex]x^5 y^5[/tex] in the binomial expansion of [tex](2x-3y)^{10}[/tex].
[tex]T_{5+1}=_{5}^{10}\textrm{C} (2x)^5\times (-3y)^{10-5}\\\\ T_{6}=_{5}^{10}\textrm{C}2^5\times (-3)^5\times x^5 y^5[/tex]
So, Coefficient of [tex]x^5 y^5[/tex]
[tex]=_{5}^{10}\textrm{C}2^5\times (-3)^5\\\\= \frac{10!}{(10-5)!\times 5!}\times 32 \times (-243)\\\\=252 \times 32 \times (-243)\\\\= -1959552[/tex]
Answer: -1959552
Step-by-step explanation:
The (r+1) th term of binomial expansion [tex](a+b)^n[/tex] is
[tex]T_{r+1}=^nC_r\ a^{n-r}\ (b)^r[/tex]
Give expression : [tex](2x-3y)^{10}[/tex]
here , a= 2x , b= -3y and n= 10
Comparing [tex]x^5y^5[/tex] to [tex] a^{n-r}\ (b)^r[/tex] , we get
r=5
The (5+1)th term will be :[tex]^{10}C_5\ (2x)^{10-5}\ (-3y)^5[/tex]
[tex]\dfrac{10!}{5!(10-5)!}\ (2x)^{5}\ (-3y)^5[/tex]
[tex]252\ (32x^5)\ (-243y^5)=252\times32\times-243x^5y^5=-1959552x^5y^5[/tex]
Hence, the coefficient of [tex]x^5y^5[/tex] =-1959552
