A) Write down the binomial expansion of (1-y)6, simplify all terms. b) By putting y = (x − x2) in (a), deduce the expansion of (1 - x + x2)6 in ascending powers as far as the term in x5. c) By taking x = 0.1, in (b), evaluate 0.916 correct to two decimal places.

$$\binom{6}{0} = 1$$ $$\binom{6}{1} = 6$$ $$\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$$ $$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ $$\binom{6}{4} = 15$$ $$\binom{6}{5} = 6$$ $$\binom{6}{6} = 1$$

Substitute these values:

$$(1-y)^6 = 1(1)(1) + 6(1)(-y) + 15(1)(y^2) + 20(1)(-y^3) + 15(1)(y^4) + 6(1)(-y^5) + 1(1)(y^6)$$ $$(1-y)^6 = 1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6$$

The binomial expansion is $\boxed{1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6}$.

Q7 b)
To deduce the expansion of $(1 - x + x^2)^6$ in ascending powers of $x$ as far as the term in $x^5$, we substitute $y = (x - x^2)$ into the expansion from part (a).

$$(1 - (x - x^2))^6 = (1 - x + x^2)^6$$

We need to expand each term of $1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6$ up to $x^5$.

$$y = x - x^2$$ $$y^2 = (x - x^2)^2 = x^2 - 2x^3 + x^4$$ $$y^3 = (x - x^2)^3 = x^3 - 3x^2(x^2) + 3x(x^2)^2 - (x^2)^3 = x^3 - 3x^4 + 3x^5 - x^6$$ $$y^4 = (x - x^2)^4 = x^4(1-x)^4 = x^4(1 - 4x + \dots) = x^4 - 4x^5 + \dots$$ $$y^5 = (x - x^2)^5 = x^5(1-x)^5 = x^5(1 - \dots) = x^5 - \dots$$

Terms $y^6$ and higher will start with $x^6$ or higher, so they are not needed for terms up to $x^5$.

$$(1 - x + x^2)^6 = 1 - 6(x - x^2) + 15(x^2 - 2x^3 + x^4) - 20(x^3 - 3x^4 + 3x^5) + 15(x^4 - 4x^5) - 6(x^5) + \dots$$ $$= 1 - 6x + 6x^2$$ $$\quad + 15x^2 - 30x^3 + 15x^4$$ $$\quad - 20x^3 + 60x^4 - 60x^5$$ $$\quad + 15x^4 - 60x^5$$ $$\quad - 6x^5$$

Constant term: 1
Coefficient of $x$: $-6$
Coefficient of $x^2$: $6 + 15 = 21$
Coefficient of $x^3$: $-30 - 20 = -50$
Coefficient of $x^4$: $15 + 60 + 15 = 90$
Coefficient of $x^5$: $-60 - 60 - 6 = -126$

The expansion of $(1 - x + x^2)^6$ up to the term in $x^5$ is:

$$\boxed{1 - 6x + 21x^2 - 50x^3 + 90x^4 - 126x^5}$$

Q7 c)
To evaluate $0.91^6$ correct to two decimal places by taking $x=0.1$ in (b).

Substitute $x=0.1$ into $(1 - x + x^2)$:

$$1 - 0.1 + (0.1)^2 = 1 - 0.1 + 0.01 = 0.9 + 0.01 = 0.91$$

So, $(1 - x + x^2)^6 = (0.91)^6$.

$$(0.91)^6 \approx 1 - 6(0.1) + 21(0.1)^2 - 50(0.1)^3 + 90(0.1)^4 - 126(0.1)^5$$ $$= 1 - 0.6 + 21(0.01) - 50(0.001) + 90(0.0001) - 126(0.00001)$$ $$= 1 - 0.6 + 0.21 - 0.05 + 0.009 - 0.00126$$

$$1 - 0.6 = 0.4$$ $$0.4 + 0.21 = 0.61$$ $$0.61 - 0.05 = 0.56$$ $$0.56 + 0.009 = 0.569$$ $$0.569 - 0.00126 = 0.56774$$

$$0.56774 \approx 0.57$$

The value of $0.91^6$ correct to two decimal places is $\boxed{0.57}$.