Therefore, the 3rd term in the expansion of (x – 10z)’ is as follows. 7! 7C2x7 - 2(-10z)? 5! (-10z)? 7:6·5!x5(-10)²(z)² 5! 2! (100z²) 7:6 2! 7:6,5(100z²) 1· 2 |× x°(100z²)
Therefore, the 3rd term in the expansion of (x – 10z)’ is as follows. 7! 7C2x7 - 2(-10z)? 5! (-10z)? 7:6·5!x5(-10)²(z)² 5! 2! (100z²) 7:6 2! 7:6,5(100z²) 1· 2 |× x°(100z²)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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8.5 q.4 Please answer and fill in each box with a red x mark, thank you
![Find the specified nth term in the expansion of the binomial.
(x – 10z)', n = 3
Step 1
According to the Binomial Theorem,
(a + b)" = an + „C1a^ - 'b + ... + „C,a" - 'b" + ... + „Cn – 1abn – 1
+ b"
the given expression is already in the form (a + b)", where a = x, b = |-10z
-10:
and
n =
7
Step 2
Now use the Binomial Theorem to find the 3rd term in the expansion of (x) + (-10z))'. Note that the rth
n!
term is „C, - 1an - (r - 1)b" - 1 and not „C,a" - "b". Also recall that „C,
and that
r! (n – r)!
(ab)" = a" b".
Therefore, the 3rd term in the expansion of (x – 10z)' is as follows.
7!
7C2x7 - 2(-10z)?
x5(-10z)²
5!
5!x5(-10)²(z)?
5! 2!
7:6,5(100z?)
2!
7.6,5(100z²)
1: 2
%3D
|× x°(100z²)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b613516-9e12-4bba-bae7-579a0cfaa601%2Fd51fb990-0f7a-402f-944f-7d676cbceb61%2Fvjys3n9_processed.png&w=3840&q=75)
Transcribed Image Text:Find the specified nth term in the expansion of the binomial.
(x – 10z)', n = 3
Step 1
According to the Binomial Theorem,
(a + b)" = an + „C1a^ - 'b + ... + „C,a" - 'b" + ... + „Cn – 1abn – 1
+ b"
the given expression is already in the form (a + b)", where a = x, b = |-10z
-10:
and
n =
7
Step 2
Now use the Binomial Theorem to find the 3rd term in the expansion of (x) + (-10z))'. Note that the rth
n!
term is „C, - 1an - (r - 1)b" - 1 and not „C,a" - "b". Also recall that „C,
and that
r! (n – r)!
(ab)" = a" b".
Therefore, the 3rd term in the expansion of (x – 10z)' is as follows.
7!
7C2x7 - 2(-10z)?
x5(-10z)²
5!
5!x5(-10)²(z)?
5! 2!
7:6,5(100z?)
2!
7.6,5(100z²)
1: 2
%3D
|× x°(100z²)
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