Asked by Jaden Snell on Jun 04, 2024
Verified
Find the 666 th term in the expansion of (2x+y) 8( 2 x + y ) ^ { 8 }(2x+y) 8 .
A) 28x3y628 x ^ { 3 } y ^ { 6 }28x3y6
B) 224x2y6224 x ^ { 2 } y ^ { 6 }224x2y6
C) 28x2y628 x ^ { 2 } y ^ { 6 }28x2y6
D) 448x3y5448 x ^ { 3 } y ^ { 5 }448x3y5
E) 112x3y5112 x ^ { 3 } y ^ { 5 }112x3y5
Expansion
The process of expressing something in an extended form, often used in mathematics to denote the development of an algebraic expression or series.
N Th Term
A term in a sequence or series that is specified by its position (n), often expressed as a formula in terms of n.
- Apply principles from the binomial theorem and Pascal's triangle to solve questions in binomial expansion.
Verified Answer
PS
Paramveer SinghJun 11, 2024
Final Answer :
D
Explanation :
The 666 th term in the expansion of (2x+y)8(2x + y)^8(2x+y)8 can be found using the binomial theorem formula for the nnn th term: Tn+1=(nr)an−rbrT_{n+1} = \binom{n}{r}a^{n-r}b^rTn+1=(rn)an−rbr , where nnn is the power of the binomial, rrr is the term number minus 1, aaa and bbb are the terms of the binomial. For the 666 th term, r=5r = 5r=5 , so we calculate T6=(85)(2x)8−5y5=56⋅8x3y5=448x3y5T_6 = \binom{8}{5}(2x)^{8-5}y^5 = 56 \cdot 8x^3y^5 = 448x^3y^5T6=(58)(2x)8−5y5=56⋅8x3y5=448x3y5 .
Learning Objectives
- Apply principles from the binomial theorem and Pascal's triangle to solve questions in binomial expansion.