Question

if the coeffecient of y in the expansion of (y^2 +k/y)^5 is 270.find k

Answer 1

http://www.wolframalpha.com/input/?i=+%28270%5E2+%2Bk%2F270%29%5E5+&dataset=

Answer 2

thanks but i dont think thats the solution for my question

Answer 3

pls help with the correct solution

Answer 4

substitute y as 270

Answer 5

http://www.wolframalpha.com/input/?i=+%28y%5E2+%2Bk%2Fy%29%5E5+
so notice that coeffecient beside y is 10k^3
so it is equal to 10k^3=270. Solving for k we get k=3

Answer 6

@ygmmumbai hope this is enough

Answer 7

so sorry myko do not understand .do you want me to get in that link ? how did you get the coefficient beside y as 10k^3?...according to the theorem we have nCr * a^n-r * b^r

Answer 8

the expansion is:
\[y^{10} + 5 y^8 (\frac{k}{y})+ 10y^6 (\frac{k}{y})^2 +10y^4 (\frac{k}{y})^3+5y^2 (\frac{k}{y})^4 + (\frac{k}{y})^5\]
The 4th term contains the "y", with coefficient of 10k^3
solve for k as @myko showed above

Answer 9

thanks so muc ..but my teacher wants it in the theorem method of nCr * a^n-r * b^r..can you guide like that please.

Answer 10

?? thats what i just did.... i used the binomial thm
nCr are coefficients 1,5,10,10,5,1
a^n-r is (y^2)^n-r
b^r is (k/y)^r

Answer 11

thanks again !!! but how do we solve for r in this theorem?

Answer 12

r varies from 0 to 5, the exponent of binomial is 5
this is how you get each term in the expansion, you plug in r=0,1,2,3,4,5

http://en.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem

Answer 13

sincere thanks to you sir! god bless

Answer 14

calculate by binomial theorem 95*405/5.25^2