The overall term \[Tr+1\] in the binomial extension is given by \[Tr+1\text{ }=\text{ }nCr\text{ }an-r\text{ }br\]

Here\[a\text{ }=\text{ }1\] , \[b\text{ }=\text{ }x\] and \[n\text{ }=\text{ }m\]

Putting the worth

\[Tr+1\text{ }=\text{ }m\text{ }Cr\text{ }1m-r\text{ }xr\]

\[=\text{ }m\text{ }Cr\text{ }xr\]

We need coefficient of \[x2\]

∴ putting \[r\text{ }=2\]

\[T2+1\text{ }=\text{ }mC2\text{ }x2\]

The coefficient of \[x2\text{ }=\text{ }mC2\]

Considering that coefficient of \[x2\text{ }=\text{ }mC2\text{ }=\text{ }6\]

\[\Rightarrow \left( m\text{ }\text{ }1 \right)\text{ }=\text{ }12\]

\[\Rightarrow m2m\text{ }\text{ }12\text{ }=0\]

\[\Rightarrow m24m\text{ }+\text{ }3m\text{ }\text{ }12\text{ }=0\]

\[\Rightarrow \left( m\text{ }\text{ }4 \right)\text{ }+\text{ }3\text{ }\left( m\text{ }\text{ }4 \right)\text{ }=\text{ }0\]

\[\Rightarrow \left( m+3 \right)\text{ }\left( m\text{ }\text{ }4 \right)\text{ }=\text{ }0\]

\[\Rightarrow m\text{ }=\text{ }\text{ }3,\text{ }4\]

We need positive worth of m so \[m\text{ }=\text{ }4\]