Help with hard binomial Expansion Question .. Please help

[kumar66]

This is really confusing don't know to go about it

Given that (1 + (2/3)x)^n (3+nx)^2=9+84x+... find value of "n"

 

[HallsofIvy]

I would presume you know that, for any n, the first two terms of \(\displaystyle (a+ bx)^n\) are \(\displaystyle a^n\) and \(\displaystyle na^{n-1}bx\). So you can write \(\displaystyle (1+ (2/3)x)^n= 1+ (2n/3)x+ \cdot\cdot\cdot\) and \(\displaystyle (3+ nx)^2= 9+ 6nx+ \cdot\cdot\cdot\) (actually, in this case "\(\displaystyle \cdot\cdot\cdot\)" is the single terms \(\displaystyle n^2x^2\)).

We only need to write the first two terms because in multiplying \(\displaystyle (a+ bx+ \cdot\cdot\cdot)(c+ dx+ \cdot\cdot\cdot)\) the first term is just ac and the next term is (ad+ bc)x. Terms of higher power in x will give higher powers in the product.

So \(\displaystyle (1+ (2/3)x)^n(3+ nx)^2= (1+ (2n/3)x+ \cdot\cdot\cdot)(9+ 6nx+ \cdot\cdot\cdot)\)

 

[kumar66]

Help with binomial Expansion

I would presume you know that, for any n, the first two terms of \(\displaystyle (a+ bx)^n\) are \(\displaystyle a^n\) and \(\displaystyle na^{n-1}bx\). So you can write \(\displaystyle (1+ (2/3)x)^n= 1+ (2n/3)x+ \cdot\cdot\cdot\) and \(\displaystyle (3+ nx)^2= 9+ 6nx+ \cdot\cdot\cdot\) (actually, in this case "\(\displaystyle \cdot\cdot\cdot\)" is the single terms \(\displaystyle n^2x^2\)).

We only need to write the first two terms because in multiplying \(\displaystyle (a+ bx+ \cdot\cdot\cdot)(c+ dx+ \cdot\cdot\cdot)\) the first term is just ac and the next term is (ad+ bc)x. Terms of higher power in x will give higher powers in the product.

So \(\displaystyle (1+ (2/3)x)^n(3+ nx)^2= (1+ (2n/3)x+ \cdot\cdot\cdot)(9+ 6nx+ \cdot\cdot\cdot)\)

I understand that (a +bx)^n the first 2 terms are a^n and na^n-1 bx but isn't na^n-1 bx in this case 2/3nx ^n-1 ?

 

[kumar66]

I understand that (a +bx)^n the first 2 terms are a^n and na^n-1 bx but isn't na^n-1 bx in this case 2/3nx ^n-1 ?

I get it "n"= 7

Thank you ever so much

Cheers

 

[Deleted member 4993]

I would presume you know that, for any n, the first two terms of \(\displaystyle (a+ bx)^n\) are \(\displaystyle a^n\) and \(\displaystyle na^{n-1}bx\). So you can write \(\displaystyle (1+ (2/3)x)^n= 1+ (2n/3)x+ \cdot\cdot\cdot\) and \(\displaystyle (3+ nx)^2= 9+ 6nx+ \cdot\cdot\cdot\) (actually, in this case "\(\displaystyle \cdot\cdot\cdot\)" is the single terms \(\displaystyle n^2x^2\)).

We only need to write the first two terms because in multiplying \(\displaystyle (a+ bx+ \cdot\cdot\cdot)(c+ dx+ \cdot\cdot\cdot)\) the first term is just ac and the next term is (ad+ bc)x. Terms of higher power in x will give higher powers in the product.

So \(\displaystyle (1+ (2/3)x)^n(3+ nx)^2= (1+ (2n/3)x+ \cdot\cdot\cdot)(9+ 6nx+ \cdot\cdot\cdot)\)

You are correct Kumar.

\(\displaystyle (1+ (2/3)x)^n(3+ nx)^2= (1+ (2n/3)x+ \cdot\cdot\cdot)(9+ 6nx+ \cdot\cdot\cdot) \ \)

\(\displaystyle = 9 + 6nx + (2/3)*n*x*9 \ + \ 4n^2x^2...... \)

\(\displaystyle = 9 + 12nx \ + \ 4n^2x^2...... \)

Then

84 (x) = 12 (nx) → n = 7