The sum of all rational terms in the expansion of (1+21/2+31/2)6 is equal to
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Step-by-Step Solution
Step 1: Identify the general term
We begin by writing the general term of the multinomial expansion (a+b+c)n. For the given expression (1+21/2+31/2)6, the general term is given by the formula shown, where r1+r2+r3=6 and r1,r2,r3 are non-negative integers.
Step 2: Simplify the general term for rationality
For a term to be rational, the powers of 2 and 3 must be integers. This means that r2/2 and r3/2 must be integers. Therefore, r2 and r3 must both be even integers.
Step 3: List possible values for r2 and r3
Since r2 and r3 must be even and their sum with r1 cannot exceed 6, the possible even values for r2 and r3 are 0, 2, 4, and 6.
Step 4: Calculate terms for valid (r1,r2,r3) combinations
We systematically list all combinations of even r2 and r3 such that r1=6−r2−r3 is non-negative. For each valid combination, we calculate the corresponding term using the simplified general term formula.
Step 5: Sum all rational terms
Finally, we sum all the calculated rational terms to find the total sum of all rational terms in the expansion.