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To expand the summation, substitute successive integer values for the index $k$ into the general term. For $k=0$, the term simplifies to $1$.

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

To expand the summation, substitute successive integer values for the index $k$ into the general term. For $k=0$, the term simplifies to $1$.

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

The $k=0$ term equals $1$ because substituting $k=0$ into the general term yields $\frac{(-1)^0 x^0}{0!}$. The video uses the standard conventions that $(-1)^0 = 1$, $x^0 = 1$, and $0! = 1$.

Conditions: The index value is $k=0$.; The standard mathematical conventions $x^0=1$ and $0!=1$ are assumed.

The $k=0$ term equals $1$ because substituting $k=0$ into the general term yields $\frac{(-1)^0 x^0}{0!}$. The video uses the standard conventions that $(-1)^0 = 1$, $x^0 = 1$, and $0! = 1$.

Conditions: The index value is $k=0$.; The standard mathematical conventions $x^0=1$ and $0!=1$ are assumed.