1) Without fully expanding (4a-b)7, determine the fifth term.
2) Find the middle term of (2c - 12π)10
1:(4aβb)7Theββfifthββtermββ(countingββfromββ0thββterm)ββisββC75(4a)5(βb)7β5=21504a5b22:(2cβ12d)10Theββmiddleββtermββisββforββ5thββpowers:C105(2c)5(β12d)5=β2006581248c5d51:\\\left( 4a-b \right) ^7\\The\,\,fifth\,\,term\,\,\left( counting\,\,from\,\,0th\,\,term \right) \,\,is\,\,\\C_{7}^{5}\left( 4a \right) ^5\left( -b \right) ^{7-5}=21504a^5b^2\\2:\\\left( 2c-12d \right) ^{10}\\The\,\,middle\,\,term\,\,is\,\,for\,\,5th\,\,powers:\\C_{10}^{5}\left( 2c \right) ^5\left( -12d \right) ^5=-2006581248c^5d^51:(4aβb)7Thefifthterm(countingfrom0thterm)isC75β(4a)5(βb)7β5=21504a5b22:(2cβ12d)10Themiddletermisfor5thpowers:C105β(2c)5(β12d)5=β2006581248c5d5
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