This is a fun question. Perhaps you will like to do the question.
Question
Using Binomial Theorem, obtain the first 3 terms in the expansion
(1-x)/[square root(4-x)]
Show that if x = 2/5, then (1-x)/[square root(4-x)] = 1/(square root 10).
Hence, calculate the value of 1/(square root 10) correct to 3 significant figures.
Thank you for your kind attention.
Regards,
ahm97sic
u dunno how do rite. so say fun question and make us do.
Hi Daisuke-kun,
I post interesting maths questions for students who like to solve maths questions.
I always post the worked out solutions to these questions.
You can see more of these interesting questions in the homework forum, addmaths.sgforums.com ie the addmaths forum and emaths.sgforums.com ie emaths forums.
Please come into my forums and try to solve these interesting questions. Please do not look at the worked solutions first when you try to do these questions. OK.
Regards,
ahm97sic
Entry is deleted
This question is fun. Perhaps you will like to do the question.
Worked Solution will be posted on Sunday ie 26th October, 2008 in the addmaths.sgforums.com ie addmaths forum.
Regards,
ahm97sic
Entry is deleted.
Can expand..? =/ No power how to expand, or do you mean must change it to
(1-x)[(4-x)^(/1/2)] then expand..
Dear iamapebble,
Yes, you need to change (1-x)/[square root(4-x)] = (1-x)[(4-x)^(-1/2)] and then
expand using binomial theorem.
Regards,
ahm97sic
PS : If you need to check your answer with my worked solution immediately, I can email it as an attached file. Please send an email address which can receive attached file.
Entry is deleted.
This is the worked solution for the fun question.
Question
Using Binomial Theorem, obtain the first 3 terms in the expansion
(1-x)/[square root(4-x)]
Show that if x = 2/5, then (1-x)/[square root(4-x)] = - 1/(square root 10).
Hence, calculate the value of - 1/(square root 10) correct to 3 significant figures.
Answer
Step 1 : Use Binomial Expansion to obtain the first three terms of the expression
(1-x)/[square root(4-x)]
= (1-x)/[square root(4(1-x/4))]
= (1/2)(1-x)/[square root(1-x/4)]
= (1/2)(1-x)(1-x/4)^(-1/2)
= (1/2)(1-x)[1 + (-1/2)(-x/4) + {(-1/2)(-3/2)}/{(1x2)}(-x/4)^2 + .......]
Use Binomial Expansion
= (1/2)(1-x)[1 + x/8 + (3/128)x^2 + .......}
= (1/2)[1+ x/8 + (3/128)x^2 - x - 1/8x^2 + ....... ]
= (1/2)[1 - (7/8)x - (13/128)x^2 + ........]
Step 2 : To show that if x = 2/5, then
(1-x)/[square root(4-x)] = 1/(square root 10)
Substitute x = 2/5 into
(1-x)/[square root(4-x)]
= (1 - 2/5)/[square root (4 - 2/5)]
= (3/5)/[square root (18/5)]
= [Square root(9/25] / [square root (18/5)] Make 3/5 into [Square root(9/25)]
= square root [(9/25)/(18/5)]
= square root [1/10]
= 1/ square root 10 (shown)
Step 3 : Use of the first three terms in the expansion to calculate the value of
1/(square root 10) correct to 3 significant figures
Substitute x = 2/5 into the first three terms in the expansion
1/square root 10 = (1/2)[1 - (7/8)x - (13/128)x^2 + ........]
= (1/2)[1 - (7/8)(2/5) - (13/128)(2/5)^2 + ........]
= 0.317 (correct to 3 significant figures)
Thank you for your kind attention.
Regards,
ahm97sic