1. The problem is to expand the expression $$(1+2y)^{-3}$$ up to the term containing $y^4$. 2. We use the Binomial series expansion for negative exponents: $$ (1+x)^n = \sum_{k=0}^\infty \binom{n}{k} x^k $$ where $$ \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} $$ and $n = -3$, $x = 2y$. 3. Calculate the first five terms (up to $k=4$): - For $k=0$: $$ \binom{-3}{0} (2y)^0 = 1 $$ - For $k=1$: $$ \binom{-3}{1} (2y)^1 = (-3)(2y) = -6y $$ - For $k=2$: $$ \binom{-3}{2} (2y)^2 = \frac{-3 \times (-4)}{2} (2y)^2 = 6 \times 4y^2 = 24y^2 $$ - For $k=3$: $$ \binom{-3}{3} (2y)^3 = \frac{-3 \times (-4) \times (-5)}{6} (2y)^3 = -10 \times 8y^3 = -80y^3 $$ - For $k=4$: $$ \binom{-3}{4} (2y)^4 = \frac{-3 \times (-4) \times (-5) \times (-6)}{24} (2y)^4 = 15 \times 16y^4 = 240y^4 $$ 4. So the expansion up to the $y^4$ term is: $$ 1 - 6y + 24y^2 - 80y^3 + 240y^4 $$ 5. This is the required expansion up to the term containing $y^4$.