What is summation from zero to n of n+r C r.1/2^r?
The expression you mentioned is a summation of a binomial coefficient multiplied by a constant value, which is raised to a power. More specifically, it is a summation of (n+r) choose r multiplied by (1/2)^r, where the summation is from zero to n.
To calculate this summation, you can use the concept of binomial coefficients and the properties of summations. Here are the steps to compute it:
Step 1: Expand the binomial coefficient using its formula: (n+r) choose r = (n+r)! / (r!(n+r-r)!).
Step 2: Simplify the expression by canceling out the factorials in the denominator of the binomial coefficient.
Step 3: Write out the simplified expression of the summation by applying the properties of summations.
Step 4: Evaluate the summation by substituting the values of r from 0 to n and adding up the terms.
Let's go through these steps in more detail.
Step 1: Expand the binomial coefficient
First, recall the formula for the binomial coefficient:
(n+r) choose r = (n+r)! / (r!(n+r-r)!).
In this case, the expression is (n+r) choose r, which can be expanded as:
(n+r) choose r = (n+r)! / (r!(n)!).
Step 2: Simplify the expression
Cancel out the (n) factorial terms:
(n+r) choose r = (n+r)(n+r-1)(n+r-2)...(n+1) / (r!(n)!)
Step 3: Write out the simplified expression of the summation
The expression can be written as a summation:
Summation from r = 0 to n of: [(n+r)(n+r-1)(n+r-2)...(n+1) / (r!(n)!) * (1/2)^r]
Step 4: Evaluate the summation
Now, substitute the values of r from 0 to n into the expression and add up the terms:
[(n+0)(n+0-1)...(n+1) / (0!(n)!) * (1/2)^0] + [(n+1)(n+1-1)...(n+1) / (1!(n)!) * (1/2)^1] + ... + [(n+n)(n+n-1)...(n+1) / (n!(n)!) * (1/2)^n]
Evaluate each term and sum them all up to get the final result.
Note: Depending on the specific values of n and r, the expression may simplify further or require additional mathematical techniques to evaluate.
I hope this explanation helps you understand how to calculate the given summation.