# Solving in different ways

**Question:**

What is the value of !!{::}^kC_k+{::}^(k+1)C_k+..+{::}^nC_k!! is:

- !!{::}^(n+1)C_k!!
- !!{::}^(n+1)C_(k+1)!!
- !!{::}^(n+1)C_(k-1)!!
- None

**Solution**

We have obviously have hacks to solve a question, but it is always better to solve a question properly when practing.

First straight forward approach:

People generally say that we can replace !!{::}^kC_k!! with !!{::}^(k+1)C_(k+1)!!, but ask them why did they think of this approach, many won't reply anything.

If you look here we are trying to add !!{::}^kC_k+{::}^(k+1)C_k!!

!!=1+((k+1)!)/(k!)!!

!!=1/(k!) (k!+(k+1)!)!!

!!=(k!)/(k!) (k+2)!!

!!=(k+2)={::}^(k+2)C_(k+1)!!

From this you could see the trend which is

!!{::}^(k+1)C_(k+1)+{::}^(k+1)C_k={::}^(k+2)C_(k+1)!!

!!{::}^(k+2)C_(k+1)+{::}^(k+2)C_k={::}^(k+3)C_(k+1)!!

!!{::}^(k+3)C_(k+1)+{::}^(k+3)C_k={::}^(k+4)C_(k+1)!!

This is the reason we substitute !!{::}^kC_k={::}^(k+1)C_(k+1)!!

So from this is clear that the answer will be !!{::}^(n+1)C_(k+1)!!

Induction approach:

We use an induction on variable !!n!!. Here we find the function which is satisfied for !!n=k!! and then prove that if this equation is satisfied for some !!n!!, then it is true for !!n+1!! also.

Coefficients approach:

If you look closely, you can see that !!{::}^rC_k!! is the coefficient of !!x^k!! in !!(1+x)^k!!.

So the given sum is equal to the sum of coefficients of !!x^k!! in !!(1+x)^k+(1+x)^(k+1)+...+(1+x)^n!!

This is the sum of coefficients of !!x^k!! in !!=1+(1+x)+(1+x)^2+.....+(1+x)^n!!

This is the coefficients of !!x^k!! in !!=((1+x)^(k+1)-1)/x!!

This is the coefficients of !!x^(k+1)!! in !!=(1+x)^(k+1)!!

!!={::}^(n+1)C_(k+1)!!