The binomial theorem practice questions — IB Maths AA

The sum of coefficients of a polynomial $P(x)$ is $P(1)$. So $\sum_{k=0}^{10}\binom{10}{k} = (1 + 1)^{10} = 2^{10} = 1024$.

The general term is $$T_k = \binom{5}{k}(2x)^{5-k}(3)^{k}.$$ For the $x^3$ term, set the exponent of $x$ equal to $3$: $$5 - k = 3 \implies k = 2.$$ Substitute $k = 2$: $$\binom{5}{2}(2x)^{3}(3)^{2} = 10 \cdot 8x^3 \cdot 9.$$ So the coefficient is $$10 \cdot 8 \cdot 9 = 720.$$

With $a = 2x$ and $b = 3$: $T_{k+1} = \binom{n}{k}(2x)^{n-k} 3^k$.

The general term in the expansion is $$T_{k+1} = \binom{5}{k}(2x)^{5-k}(3)^{k}$$ The power of $x$ is $5-k$, so set $$5 - k = 3 \implies k = 2$$ Substituting $k = 2$: $$\binom{5}{2}(2)^{3}(3)^{2} = 10 \cdot 8 \cdot 9 = 720$$

The general term in the expansion is $$\binom{5}{r}(2)^{5-r}(3x)^{r}.$$

The term in $x^3$ comes from $r = 3$: $$\binom{5}{3}(2)^{5-3}(3)^{3}\,x^{3}.$$

Evaluate each factor: $$\binom{5}{3} = 10,\quad 2^{2} = 4,\quad 3^{3} = 27.$$

Multiply: $$10 \times 4 \times 27 = 1080.$$

So the coefficient of $x^3$ is $1080$.

$5 - k = 3 \Rightarrow k = 2$. Coefficient: $\binom{5}{2} \cdot 2^2 = 10 \cdot 4 = 40$.

$\sum_{k=0}^{6}(-1)^k \binom{6}{k} = (1 - 1)^6 = 0$.

The coefficient of $x^2$ corresponds to $k = 2$: $\binom{4}{2} = 6$.

$(x - 1)^3 = x^3 - 3x^2 + 3x - 1$.

The binomial formula gives $(a + b)^2 = a^2 + 2ab + b^2$.