Question

The coefficient of x
3
in the expansion of (a + x)
5 + (2 − x)
6
is 90. Find the value of the positive constant a.

Answer 1

it is coefficitent of x^3 and in the expansion of (a + x )^5 + (2-x ) ^ 6

Answer 2

can you tell me the expression for rth term in the expansion of (x + y)^n ?

Answer 3

tr= nCr.a^n-r.b^r

Answer 4

its nCr x^(n-r) y^r

Answer 5

tr = nCr-1 x^(n-r) y^r

so all you need to do is to find the coefficient of x^3 in both of the series and add them

Answer 6

\[(a+x)^{5}=a ^{5}+5a ^{4}x+10a ^{3}x ^{2}+10a ^{2}x ^{3}+5ax ^{4}+x ^{5}\]
\[(2-x)^{6}=x ^{6}-12x ^{5}+60x ^{4}-160x ^{3}+240x ^{2}-192x+64\]
Take the coefficitents near x^3 from these equations and do this:
\[10a ^{2}-160=90\]

Answer 7

in (a+x)^5 its 5Cr (a)^(5-r) x^r and in (2-x)^6 its 6Cr (2)^(6-r) x^r

now adding both we have taking only the x terms r=3 simply

5C3(a)^2 x^3 + 6C3(2)^3 x^3 = 90 (given)

so considering only the terms you remove x^3 and equate them for a


Hope u understood?

Answer 8

Yes i did. THANKYOU V.MUCH every one. i got the answer :)