(a) Expand (x + y)⁵ using the binomial theorem. (b) Use Pascal's identity to prove that the sum of the first n positive integers is equal to n(n+1)/2.

(a) Expansion of $(x+y{)}^5$ using the Binomial Theorem

The Binomial Theorem states that for any positive integer $n$, $$(x+y{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}{y}^k$$ where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is the binomial coefficient, calculated as: $$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$$

For $(x+y{)}^5$, we substitute $n=5$: $$(x+y{)}^5=\sum _{k=0}^5(\genfrac{}{}{0ex}{}{5}{k})x^{5-k}{y}^k$$

Now, we compute each term individually:

1. For $k=0$: $$\left(\genfrac{}{}{0ex}{}{5}{0}\right)x^5{y}^0=1\cdot x^5\cdot 1=x^5$$

2. For $k=1$: $$\left(\genfrac{}{}{0ex}{}{5}{1}\right)x^4{y}^1=5\cdot x^{4}y=5x^{4}y$$

3. For $k=2$: $$\left(\genfrac{}{}{0ex}{}{5}{2}\right)x^3{y}^2=10\cdot x^3{y}^2=10x^3{y}^2$$

4. For $k=3$: $$\left(\genfrac{}{}{0ex}{}{5}{3}\right)x^2{y}^3=10\cdot x^2{y}^3=10x^2{y}^3$$

5. For $k=4$: $$\left(\genfrac{}{}{0ex}{}{5}{4}\right)x^1{y}^4=5\cdot x{y}^4=5x{y}^4$$

6. For $k=5$: $$\left(\genfrac{}{}{0ex}{}{5}{5}\right)x^0{y}^5=1\cdot 1\cdot y^5=y^5$$

Combining all terms, the expansion is: $$(x+y{)}^5=\boxed{x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5}$$

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(b) Proof that the sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$ using Pascal’s Identity

Pascal’s Identity

Pascal’s Identity states that for any positive integers $n$ and $k$, $$\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)$$

Sum of the First $n$ Positive Integers

We aim to prove: $$1+2+3+\cdots +n=\frac{n(n+1)}{2}$$

Proof Using Combinatorial Interpretation

Consider the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)$. By definition, $$\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)=\frac{(n+1)\cdot n}{2}=\frac{n(n+1)}{2}$$

Now, using Pascal’s Identity for $k=2$: $$\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)=\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\left(\genfrac{}{}{0ex}{}{n}{1}\right)$$

But $\left(\genfrac{}{}{0ex}{}{n}{1}\right)=n$, and $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$. Substituting: $$\frac{n(n+1)}{2}=\frac{n(n-1)}{2}+n$$

Simplify the right-hand side: $$\frac{n(n-1)}{2}+n=\frac{n(n-1)+2n}{2}=\frac{n^2-n+2n}{2}=\frac{n^2+n}{2}=\frac{n(n+1)}{2}$$

This confirms the identity, but we need a direct connection to the sum of integers.

Alternative Proof Using Binomial Coefficients

We can express the sum $S=1+2+3+\cdots +n$ in terms of binomial coefficients. Notice that: $$\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)=\sum _{k=2}^{n+1}\left(\genfrac{}{}{0ex}{}{k}{2}\right)-\sum _{k=2}^n\left(\genfrac{}{}{0ex}{}{k}{2}\right)$$ However, a more straightforward approach is to use the Hockey Stick Identity (a special case of Pascal’s Identity), which states: $$\sum _{i=r}^n\left(\genfrac{}{}{0ex}{}{i}{r}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r+1}\right)$$

For $r=1$: $$\sum _{i=1}^n\left(\genfrac{}{}{0ex}{}{i}{1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)$$

But $\left(\genfrac{}{}{0ex}{}{i}{1}\right)=i$, so: $$1+2+3+\cdots +n=\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)=\frac{n(n+1)}{2}$$

Thus, we have proven that: $$\boxed{1+2+3+\cdots +n=\frac{n(n+1)}{2}}$$