Question

The coefficient of x^5 in the binomial expansion of (1+5x)^8 is the same as the coefficient of x^4 in the expansion of (a+5x)^7.Find the value of a

Answer 1

Let's use binomial theorem for this one.
Did you learn binomial theorem, yet?

Answer 2

\[(a+b)^n=\left(\begin{matrix}n \\ 0\end{matrix}\right)a^n b^0+\left(\begin{matrix}n \\ 1\end{matrix}\right)a^{n-1} b^1+...+\left(\begin{matrix}n \\ n\end{matrix}\right)a^0 b^n\]

Answer 3

From here, we can see that the kth term of the expansion is\[\left(\begin{matrix}n \\ k-1\end{matrix}\right) a^{n-k+1} b^{k-1}\]

Answer 4

ok

Answer 5

For the first expansion, (1+5x)^8, the term with x^5 is the fourth term.

Answer 6

y

Answer 7

Let's substitute a=1 b=5x

Answer 8

and n=8

Answer 9

\[(1+5x)^8=\left(\begin{matrix}8 \\ 0\end{matrix}\right) (1)^8(5x)^0+...+\left(\begin{matrix}8 \\ 8\end{matrix}\right)(1)^0(5x)^8\]

Answer 10

From here, the kth term of this expansion is \[\left(\begin{matrix}8 \\ k-1\end{matrix}\right)(1)^{9-k}(5x)^{k-1}\]Since the term contains x^5,\[x^{k-1}=x^5\]So k=6.
Therefore this is the 6th term.

Answer 11

And my previous answer is wrong.
Sorry. My mind must've slipped or something.

Answer 12

previous answer is wrong.
which one

Answer 13

The term containing x^5 is the 6th term, not fourth term.

Answer 14

OK

Answer 15

Now, we can find the coefficient of the 6th term.\[\left(\begin{matrix}8 \\ 6-1\end{matrix}\right)(1)^{9-6}(5x)^{5}=\left(\begin{matrix}8 \\ 5\end{matrix}\right)(5^5)x^5=(56)(5^5)x^5\]

Answer 16

I left it like that on purpose

Answer 17

You'll see why later

Answer 18

what do I need to do nw

Answer 19

Now, we do the same thing to the second expansion, but this time, we find the term with x^4

Answer 20

Can you try this on your own?

Answer 21

will try nw

Answer 22

Just to let you know.. the given answer is 2

Answer 23

Ok.

Answer 24

Is this correct ?