1. Given the identity to prove: $(n \text{ choose } r) + (n \text{ choose } r-1) = (n+1 \text{ choose } r)$. This is a key binomial coefficient identity. 2. Recall the definition of binomial coefficient: $$ (n \text{ choose } r) = \frac{n!}{r! (n-r)!} $$ 3. Express the left-hand side (LHS) with factorials: $$ \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-(r-1))!} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} $$ 4. Find common denominator for LHS: Common denominator is $r!(n-r+1)!$ because $$r! = r \times (r-1)!$$ and $$(n-r+1)! = (n-r+1)(n-r)!$$ 5. Rewrite first term: $$\frac{n!}{r!(n-r)!} = \frac{n! (n-r+1)}{r!(n-r+1)(n-r)!} = \frac{n! (n-r+1)}{r!(n-r+1)!}$$ 6. Rewrite second term multiplying numerator and denominator by $r$: $$ \frac{n!}{(r-1)!(n-r+1)!} = \frac{n! r}{r (r-1)! (n-r+1)!} = \frac{n! r}{r! (n-r+1)!} $$ 7. Add the two fractions: $$ \frac{n!(n-r+1)}{r!(n-r+1)!} + \frac{n! r}{r!(n-r+1)!} = \frac{n! (n-r+1 + r)}{r! (n-r+1)!} = \frac{n! (n+1)}{r! (n-r+1)!} $$ 8. Note $\frac{n! (n+1)}{r! (n-r+1)!} = \frac{(n+1)!}{r! ((n+1)-r)!}$ which is the definition of $(n+1 \text{ choose } r)$. 9. Thus, proven: $$(n \text{ choose } r) + (n \text{ choose } r-1) = (n+1 \text{ choose } r)$$. 10. Now let's prove by induction that: $$ (1 + x)^n = \sum_{k=0}^n (n \text{ choose } k) x^k $$ 11. Base case $n=0$: $$ (1 + x)^0 = 1 $$ and $$ \sum_{k=0}^0 (0 \text{ choose } k) x^k = (0 \text{ choose } 0) x^0 = 1 $$ True. 12. Inductive hypothesis: assume true for $n$: $$ (1 + x)^n = \sum_{k=0}^n (n \text{ choose } k) x^k $$ 13. Inductive step: prove for $n+1$: $$ (1 + x)^{n+1} = (1 + x)(1 + x)^n = (1 + x) \sum_{k=0}^n (n \text{ choose } k) x^k $$ 14. Multiply out: $$ = \sum_{k=0}^n (n \text{ choose } k) x^k + \sum_{k=0}^n (n \text{ choose } k) x^{k+1} $$ 15. Rewrite sums aligning powers of $x$: $$ = (n \text{ choose } 0) x^0 + \sum_{k=1}^n \left[(n \text{ choose } k) + (n \text{ choose } k-1)\right] x^k + (n \text{ choose } n) x^{n+1} $$ 16. Using the binomial identity from step 1: $$ (n \text{ choose } k) + (n \text{ choose } k-1) = (n+1 \text{ choose } k) $$ 17. So expression becomes: $$ = (n+1 \text{ choose } 0) x^0 + \sum_{k=1}^n (n+1 \text{ choose } k) x^k + (n+1 \text{ choose } n+1) x^{n+1} = \sum_{k=0}^{n+1} (n+1 \text{ choose } k) x^k $$ 18. This completes the inductive step and therefore the binomial theorem is proven by induction. Final answer: $$(n \text{ choose } r) + (n \text{ choose } r-1) = (n+1 \text{ choose } r)$$ and $$(1 + x)^n = \sum_{k=0}^n (n \text{ choose } k) x^k $$ proven by induction.