Refer the below solution for the proper understanding.
Step-by-step explanation:
Solution :
Using the Binomial Theorem -
[tex](a + b)^n=nc_0a^nb^0+nc_1a^n^-^1b^1+..............+nc_na^0b^n\\[/tex]
1) [tex](x+2)^6[/tex]
[tex](x+2)^6=x^6+12x^5+60x^4+160x^3+240x^2+192x+64[/tex]
2) [tex](x-4)^4[/tex]
[tex](x-4)^4=x^4-16x^3+96x^2-256x+256[/tex]
3) [tex](2x+3)^5[/tex]
[tex](2x+3)^5=32x^5+240x^4+720x^3+1080x^2+810x+243[/tex]
4) [tex](2x-3y)^4[/tex]
[tex](2x-3y)^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4[/tex]
Using Pascals triangle,
1) [tex](x+2)^6[/tex]
a = x , b =2 and n = 6 , using the seventh row of pascals triangle
[tex](x+2)^6= 1(x^6)(6^0)+6(x^5)(6^1)+15(x^4)(6^2)+20(x^3)(6^3)+15(x^2)(6^4)+6(x^1)(6^5)+1(x^0)(6^6)[/tex]
[tex](x+2)^6 = x^6+36x^5+540x^4+432x^3+19440x^2+46656x+46656[/tex]
2) [tex](x-4)^4[/tex]
a = x , b = -4 and n = 4, using the fifth row of pascals triangle
[tex](x-4)^4= x^4-16x^3+96x^2-256x+256[/tex]
3) [tex](2x+3)^5[/tex]
a = 2x , b = 3 and n = 5, using the sixth row of pascals triangle
[tex](2x+3)^5=32x^5+240x^4+720x^3+1080x^2+810x+243[/tex]
4) [tex](2x-3y)^4[/tex]
a = 2x , b = -3y and n = 4, using the fifth row of pascals triangle
[tex](2x-3y)^4= 16x^4-96x^3+216x^2-216x+81[/tex]
Now, in the expansion of
[tex](3a + 4b)^8[/tex]
The degree of the term must add up to 8.
[tex]ab^7[/tex] is of degre 8 than the other terms, this is because [tex]ab^7[/tex] is same as [tex]a^1b^7[/tex], adding the components 1 and 7, we get 8.
For more information, refer the link given below
https://brainly.com/question/16978014?referrer=searchResults
