According to the given question,

G.P: \[1\text{ }+\text{ }4\text{ }+\text{ }16\text{ }+\text{ }64\text{ }+\text{ }\ldots \ldots ..\]

Here,

\[a\text{ }=\text{ }1\text{ }and\text{ }r\text{ }=\text{ }4/1\text{ }=\text{ }4\text{ }\left( r\text{ }>\text{ }1 \right)\]

And,

\[{{S}_{n}}~=\text{ }5461\]

We know that,

\[{{S}_{n}}~=\text{ }a({{r}^{n~}}-\text{ }1)/\text{ }r\text{ }-\text{ }1\]

\[\Rightarrow {{S}_{n}}~=\text{ }\left( 1 \right)({{\left( 4 \right)}^{n~}}-\text{ }1)/\text{ }4\text{ }-\text{ }1\]

\[=\text{ }({{4}^{n}}-\text{ }1)/3\]

\[5461\text{ }=\text{ }({{4}^{n}}-\text{ }1)/3\]

\[16383\text{ }=\text{ }{{4}^{n}}-\text{ }1\]

\[{{4}^{n}}~=\text{ }16384\]

\[{{4}^{n}}~=\text{ }{{4}^{7}}\]

\[n\text{ }=\text{ }7\]

Hence, \[7\text{ }terms\]of the G.P should be added to get a sum of \[5461.\]