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1) an=a1+(n-1)d=1+(n-1)*3=3n-2bn=b1*q^(n-1)=2*2^(n-1)=2^n2) b4=2^4=16
If it is in, then 3n-2=16 3n=18 n=9 is the 9th term.
3) b6=a22=64
b8=86=256
b10=a342=1024
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bn=2^an
bn+1=2 lead stool an+1
bn+1 bn=2 an+1 an=2 an=2 (an+1-an)=2 d
bn) is a proportional series.
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The number series {an} and {bn} are both equal difference series.
Let the tolerance of the series {an} be da, and the tolerance of {bn} is db1+b1=7
a3+b3=a1+2da+b1+2db=7+2(da+db)=21da+db=7
a5+b5a1+4da+b1+4db
a1+b1+4(da+db)
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The sum of the tolerances of the two series is 7, so 28+7=35
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c1=1=a1+b1 b1=0 a1=1
c2=1=a2+b2 a1q+b1+d=1 q+d=1c3=2=a3+b3=a1q 2+b1+2d q 2+2d=2 q=2 d=-1
The sum of the first 10 terms of the sequence {cn} = the sum of the first 10 terms of the sequence {an} + the sum of the first 10 terms of the series {bn}.
The sum of the first ten terms of the number series {evoking the skin an} of the first ten terms of the chain = a1 (1-2 10) (1-2) = 2 10-1 = 511
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Wait, I'll do the math.
a3+a5=2a4=b4
i.e. 2+6d=q 3---1).
Because b2b3=a8
That is, q*q 2=1+7d
q^3=1+7d---2)
1) (2) Synopids d=1, q=2
So s10 = (1 + 10) * 10 2 = 55
t10=(2^10-1)/(2-1)=1023
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Let the tolerance be d and the tolerance be q
a2=b2a1+d=b1q
a1 = 1, b1 = 1 substitution, get.
d+1=qd=q-1
2a3-b3=1
2(a1+2d)-b1q²=1
a1=1, b1=1 substitution, arrangement, getting.
d=(q²-1)/4
q-1=(q²-1)/4
Finish, get (q-1)(q-3)=0 get q=1 or q=3(1) when q=1.
d=q-1=1-1=0
an=a1+(n-1)d=1+0·(n-1)=1bn=b1qⁿ⁻¹=1·1ⁿ⁻¹=1
The general formula for the series is an=1, and the formula for the series is bn=1(2) when q=3.
d=q-1=1-1=2
an=a1+(n-1)2=1+2· (n-1)=2n-1bn=b1q =1·3 =3 The general formula for the series is an=2n-1, and the general formula for the series is bn=3
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Let the tolerance d and the tolerance q be brought to the last two formulas respectively, and you can find them.
Trust me, that's right.
Method 1: When there are 2n terms in the equal difference series, the sum of the even terms - the sum of the odd terms = nd (i.e. n * tolerance) and: the sum of the even terms + the sum of the odd terms = the sum of the number series (i.e. the sum of the first 2n terms) So: the sum of the series = 2 * the sum of the odd terms + nd >>>More
a1 + a2 + a8 + a9 = = a3 + a4 + a6 + a7 = 4a5 so 5 a5 = 450 to get a5 = 90 >>>More
The formula for the nth term of the equal difference series an=a1+d(n-1) (a1 is the first term, d is the tolerance, and n is the number of terms). >>>More
1. an==a1+(n-1)d, then a3=a1+2d==-6, and a6==a1+5d=0, the connection equation gives a tolerance of 2, and the first term is -10, so an=2n-12 >>>More
Let the tolerance be d a3, a6, a7 into an equal proportional series. >>>More