SOLUTION: could some on plz help. Find the last three terms of (2x+y)^5
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Question 524390: could some on plz help. Find the last three terms of (2x+y)^5
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
could some on plz help.
Find the last three terms of
(2x+y)^5 = ........+5C2*(2x)^2*y^3 + 5C1*(2x)y^4 + 5C0*(2x)^0*y^5
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= ......10*4x^2*y^3 + 5*2xy^4 + y^5
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= ......40x^2y^3 + 10xy^4 + y^5
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Cheers,
Stan H.
================
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I'll do one better: I'll show you the form for general term of ^n)
^n\ =\ \sum_{k\,=\,0}^n\ \left(n\cr k\right\)a^{n-k}b^k)
Where )
is the number of combinations of 
things taken 
at a time and is calculated by !})
For your problem: 
, 
, 
and 
John
My calculator said it, I believe it, that settles it
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