91Ó°ÊÓ

The Binomial Theorem states that for any numbers a and b, and natural number n, the expression \((a + b)^n\) can be expanded in the form of \(\sum_{k=0}^{n} {n \choose k} a^{n-k} b^{k}\) where \({n \choose k}\) denotes the binomial coefficient, which can be calculated using the formula \({n \choose k} = \frac{n!}{k!(n - k)!}\).

The given expression is \((a+2 b)^{6}\). To apply the Binomial Theorem to this expression, a is equivalent to 'a', b is equivalent to '2b' and n is '6'. So, its expansion according to the theorem would be \(\sum_{k=0}^{6} {6 \choose k} a^{6-k} (2 b)^{k}\).

Now, break down the expression into separate terms. There should be 7 terms as k runs from 0 to 6 inclusive. The terms are: 1. For \(k=0\): \( {6 \choose 0} a^{6} (2 b)^{0} = a^{6} \) as \({6 \choose 0} = 1\) and \(b^{0} = 1\). 2. For \(k=1\): \( {6 \choose 1} a^{5} (2 b)^{1} = 6a^{5}2b = 12 a^{5} b \) 3. For \(k=2\): \( {6 \choose 2} a^{4} (2 b)^{2} = 15 a^{4}4b^{2} = 60 a^{4} b^{2} \)4. For \(k=3\): \( {6 \choose 3} a^{3} (2 b)^{3} = 20 a^{3}8b^{3} = 160 a^{3} b^{3} \)5. For \(k=4\): \( {6 \choose 4} a^{2} (2 b)^{4} = 15 a^{2}16b^{4} = 240 a^{2} b^{4} \)6. For \(k=5\): \( {6 \choose 5} a^{1} (2 b)^{5} = 6 a^{1}32b^{5} = 192 a^{1} b^{5} \)7. For \(k=6\): \( {6 \choose 6} a^{0} (2 b)^{6} = b^{6}64 \) as \({6 \choose 6} = 1\) and \(a^{0} = 1\)

Now we combine all terms from step 3 to get the expanded form of \((a+2 b)^{6}\) which is: \(a^{6} + 12 a^{5} b + 60 a^{4} b^{2} + 160 a^{3} b^{3} + 240 a^{2} b^{4} + 192 a^{1} b^{5} + 64 b^{6}\)