Q.Expand $(x+2)^5$ using the binomial theorem.

With $n=5$, the expansion has 6 terms, and the binomial coefficients needed are exactly the 5th row of Pascal's triangle: $1,5,10,10,5,1$.

By the Binomial Theorem, $(x+2)^5 = \sum_{r=0}^{5} \,^{5}C_{r}\,x^{5-r}(2)^{r}$. Computing each term:

• $r=0$: $^{5}C_{0}\,x^5(2)^0 = 1\cdot x^5\cdot1 = x^5$
• $r=1$: $^{5}C_{1}\,x^4(2)^1 = 5\cdot x^4\cdot2 = 10x^4$
• $r=2$: $^{5}C_{2}\,x^3(2)^2 = 10\cdot x^3\cdot4 = 40x^3$
• $r=3$: $^{5}C_{3}\,x^2(2)^3 = 10\cdot x^2\cdot8 = 80x^2$
• $r=4$: $^{5}C_{4}\,x^1(2)^4 = 5\cdot x\cdot16 = 80x$
• $r=5$: $^{5}C_{5}\,x^0(2)^5 = 1\cdot1\cdot32 = 32$

Adding all six terms: $(x+2)^5 = x^5+10x^4+40x^3+80x^2+80x+32$. …

Forgetting to raise the constant $2$ to the matching power $r$ in each term (e.g. writing $5x^4\cdot2$ instead of correctly computed $10x^4$ is fine, but a common slip is writing just $5x^4$ wit …