1. The problem is to use the Binomial Theorem to prove the expansion of $ (a+b)^n $. 2. The Binomial Theorem states that for any integer $ n \geq 0 $: $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ where $ \binom{n}{k} = \frac{n!}{k!(n-k)!} $ is the binomial coefficient. 3. This formula means that when you expand $ (a+b)^n $, the term for each $ k $ is the coefficient $ \binom{n}{k} $ multiplied by $ a $ raised to the power $ n-k $ and $ b $ raised to the power $ k $. 4. To prove this, we use mathematical induction: 5. Base case: For $ n=0 $, $$ (a+b)^0 = 1 $$ which matches the sum with $ k=0 $: $$ \binom{0}{0} a^0 b^0 = 1 $$ 6. Inductive step: Assume the theorem holds for $ n $, i.e., $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ 7. For $ n+1 $, $$ (a+b)^{n+1} = (a+b)(a+b)^n = (a+b) \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ 8. Distribute $ (a+b) $: $$ = \sum_{k=0}^n \binom{n}{k} a^{n-k+1} b^k + \sum_{k=0}^n \binom{n}{k} a^{n-k} b^{k+1} $$ 9. Reindex the second sum by letting $ j = k+1 $: $$ = \sum_{k=0}^n \binom{n}{k} a^{n+1-k} b^k + \sum_{j=1}^{n+1} \binom{n}{j-1} a^{n+1-j} b^j $$ 10. Combine the sums: $$ = a^{n+1} + \sum_{k=1}^n \left[ \binom{n}{k} + \binom{n}{k-1} \right] a^{n+1-k} b^k + b^{n+1} $$ 11. Using Pascal's identity: $$ \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} $$ 12. So the sum becomes: $$ = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k} b^k $$ 13. This completes the induction and proves the Binomial Theorem. Final answer: $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$