Q.The coefficient of $x^6$ in $(2+2x)^{10}$ is

Rather than expanding an entire binomial, most exam-style questions only need ONE specific term — the middle term, or the term containing a chosen power of $x$.

The general (or $(r+1)^{th}$) term. In the expansion of $(a+b)^n$,

This single formula is the workhorse for every "find the coefficient of $x^k$" or "find the constant term" problem: substitute the given $a,b$, simplify the power of $x$ in $T_{r+1}$ as a function of $r$, set that power equal to the target exponent (or to $0$, for a constant term), solve for $r$, and substitute back to get the actual term/coefficient.

Middle term(s).

• If $n$ is even, there is a single middle term, the $\left(\tfrac n2+1\right)^{th}$ term: $T_{\frac n2+1}={}^nC_{n/2}\,a^{n/2}b^{n/2}$.
• If $n$ is odd, there are two middle terms, the $\left(\tfrac{n-1}2+1\right)^{th}$ and $\left(\tfrac{n+1}2+1\right)^{th}$ terms. Their binomial coefficients are $^nC_{(n-1)/2}$ and $^nC_{(n+1)/2}$ — and since $(n-1)/2+(n+1)/2=n$, these are complementary coefficients, hence always equal by $^nC_r={}^nC_{n-r}$.
• The greatest coefficient in $(a+b)^n$ is $^nC_{n/2}$ if $n$ is even, and the (equal) pair $^nC_{(n-1)/2},{}^nC_{(n+1)/2}$ if $n$ is odd.

Working technique for "coefficient of $x^k$" problems. Write the general term, collect the power of $x$ into one exponent expression in $r$, and solve. If a product of two binomial expansions is involved (e.g. $(1+x^3)^{50}(x^2+\tfrac1x)^5$), take the general term of each factor separately, add the exponents, and sum the coefficient contributions over every valid pair of indices that hits the target power.

Ratio-of-terms problems. When told binomial coefficients of consecutive terms are in a given ratio (e.g. $1:7:42$) or in AP, use $\dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{n-r+1}{r}$ to convert each ratio condition into a linear equation in $n$ and $r$, then solve the resulting system.