How do you expand the summation $\sum_{k=0}^{\infty}\frac{(-1)^k x^{2k}}{(2k)!}$ term by term?

Conditions

• The summation index $k$ starts at $0$ and increases by $1$ for each term.
• The general term is $\frac{(-1)^k x^{2k}}{(2k)!}$.

Reasoning, step by step

1. Substitute $k=0$ into the general term to get $\frac{(-1)^0 x^0}{0!} = 1$.
2. Substitute $k=1$ to get $\frac{(-1)^1 x^2}{2!} = -\frac{x^2}{2!}$.
3. Substitute $k=2$ to get $\frac{(-1)^2 x^4}{4!} = +\frac{x^4}{4!}$.
4. Substitute $k=3$ to get $\frac{(-1)^3 x^6}{6!} = -\frac{x^6}{6!}$.
5. Write the terms in order and append an ellipsis to indicate the pattern continues.

Example

The video explicitly walks through substituting $k=0, 1, 2,$ and $3$ into the formula, showing how each value produces one term in the expanded series.

Common misconceptions

• Assuming the index starts at $1$ instead of $0$.
• Forgetting that $(-1)^k$ alternates the sign of each term.
• Misapplying the factorial in the denominator, such as using $k!$ instead of $(2k)!$.

Watch the explanation

Explore next

• Why does the $k=0$ term of this series equal $1$?
• What does the trailing ellipsis mean in the expanded series?
• How do you calculate the partial sums of this series?