Binomial Coefficient

[Mampac]

hey everyone,
please help.

i don't even know how to start

 

[Steven G]

You have to try something! Did you try to expand the lhs? Can you find 63 factors that are greater than 25?

 

[JeffM]

Thank you for the exact statement of the problem. In this case, it would have helped had you said what you were studying currently. I am going to guess that it is the binomial expansion and provide a BIG hint.

[MATH](a + b)^m = \sum_{j=0}^m \dbinom{m}{j} * a^j * b^{(m-j)}[/MATH].

Have you ever seen that?

There is a special case

[MATH](1 + b)^m = \sum_{j=0}^m \dbinom{m}{j} * 1^j * b^{(m-j)} = \sum_{j=0}^m \dbinom{m}{j} * b^{(m-j)} = \sum_{k=0}^m \dbinom{m}{k} * b^k.[/MATH]
Can you justify EACH of those steps?

Now do you see something in your problem that looks like that final expression?

 

[Steven G]

Thank you for the exact statement of the problem. In this case, it would have helped had you said what you were studying currently. I am going to guess that it is the binomial expansion and provide a BIG hint.

[MATH](a + b)^m = \sum_{j=0}^m \dbinom{m}{j} * a^j * b^{(m-j)}[/MATH].

Have you ever seen that?

There is a special case

[MATH](1 + b)^m = \sum_{j=0}^m \dbinom{m}{j} * 1^j * b^{(m-j)} = \sum_{j=0}^m \dbinom{m}{j} * b^{(m-j)} = \sum_{k=0}^m \dbinom{m}{k} * b^k.[/MATH]
Can you justify EACH of those steps?

Now do you see something in your problem that looks like that final expression?

I was thinking (1+1)^n would help but quickly ruled that out. Why did I not then think of (1+b)^n? I wish I could think better!

 

[JeffM]

I was thinking (1+1)^n would help but quickly ruled that out. Why did I not then think of (1+b)^n? I wish I could think better!

Most days you have the insight. On a rare day, I do.