Example: 7  
Find the sum given by 
Solution: 7  
We have to plan an approach wherein we are able to generate with . We can generate one with every , as we did earlier, and which is now repeated here:
Differentiating both sides with respect to , we have
Now we have reached the stage where we have an with every . We need to think how to get the other . If we differentiate once again, we’ll have with every instead of (understand this point carefully). To ‘makeup’ for the power that falls one short of the required value, we simply multiply by on both sides of the relation above to obtain:
It should be evident now that the next step is differentiation:
Now we simply substitute to obtain
The required sum is thus

Example: 8  
Evaluate the following sums:

Solution: 8(a)  
The first sum contains only the evennumbered binomial coefficients, while the second contains only oddnumbered ones. Recall that we have already evaluated the sum given by
Note that is the sum of and , i.e.,
Thus, if we determine , is automatically determined, and viceversa. Let us try to determine first.
Consider again the general expansion
Integrating with respect to , we have (we have not yet decided the limits)
Since we are trying to determine which contains only the evennumbered terms, we have to choose the limits of integration such that the oddnumbered terms vanish. This is easily achievable by setting and (understand this carefully). Thus, we have
which implies that

Solution: 8(b)  
is now simply given by 