- If (1+2x+3x2)10=a0+a1x+a2x2+.......+a20x20 then find the coefficients of a1 ; a2 ; a3 and a20
Solution
From our hypothesis we already had:
(1+2x+3x2)10=a0+a1x+a2x2+.......+a20x20
We will find the coefficients of a1 ; a2 ; a3 and a20
Which meant they are the coefficients of x ; x2 ; x3 and x20 .
We can rewrite (1+2x+3x2)10=[1+x(2+3x)]10
We will use the Newton's Formula
(a+b)n=n∑i=0C(n,i)an-i.bi
Then, [1+x(2+3x)]10=C(10,0)+C(10,1)x(2+3x)+C(10,2)x2.(2+3x)2
+C(10,3)x3.(2+3x)3+....+C(10,10)x10.(2+3x)10
We observe that:
Coef(x)=2C(10,1)=20 then, a1=20
Coef(x2)=22C(10,2)=4.45=180 then, a2=180
Coef(x3)=12C(10,2)+8C(10,3)=45.12+120.8=1500 then a3=1500
Coef(x20)=C(10,10).310=310 then, a20=310
2. If (1+x+2x2)20=a0+a1x+a2x2+......+a40x40 Let's find the value of: a0+a2+a4+-----+a38
Solution
We already had: (1+x+2x2)20=a0+a1x+a2x2+......+a40x40
If x=1 then, a0+a1+a2+.....+a40=420
If x=-1 then a0-a1+a2-a3+......-a39+a40=220
Sum side to side: 2a0+2a2+2a4+.............+2a40=420+220
Then. a0+a2+a4+.......+a40=420+2202
Therefore, a0+a2+a4+......+a38=420+2202-a40
=240+2202-220
=240+2202
Hence, a0+a2+a4+.....+a38=240+2202
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