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okay so using Binomial Theorem determine the 5th term of (x+2b)^4 without fully expanding.
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find the 5th term of \[ {(x+2b)}^4\] without fully expanding?
\[C_4^4(x)^{0}(2b)^4=16b^4\]
thats a nice way of getting around the nCr binom latex stuff
use this \[(a+b)^n=\sum_{k=0}^{n}C_n^k a^{n-k}b^k\] there \[C_n^k=\frac{n!}{(n-k)!k!}, k!=k \times (k-1)\times (k-2)\dots 2\times1\]
^nC_r works too i think
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\[^n C_r\]
I like \( \binom{n}{r} \) Just to add @amistre64 lol
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