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Anonymous users2024-02-09According to the first n terms and formulas.
sn=na1+n(n-1)*d/2
s2n=2na1+2n(2n-1)*d 2 is obtained by dividing the two equations into a quadratic equation with n as the root.
sn=(d/2)n^2+(a1-d/2)ns2n=(2d)n^2+(2a1-d)n
sn s2n is a constant, i.e., the root coefficient (-b 2a) of the above two equations is the same.
The column formula is simplified to obtain (2a1-d)d=(2a1-d)*2d, and d1=0;d2=2a1
d1=0 an=a1
d2=2a1 an=na1+(n-1)d=3na1-2a1 took me an hour and a half to check Don't forget.
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Anonymous users2024-02-08sn=n(a1+an)/2=n(2a1+(n-1)d)/2s2n=2n(2a1+(2n-1)d)/2sn=n(a1+an)/2=n(2a1+(n-1)d)/2s2n=2n(2a1+(2n-1)d)/2sn/s2n=(2a1+(n-1)d)/2(2a1+(2n-1)d)＝k
The only way to make sn s2n unrelated to n is to d=0 so an=a1
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Anonymous users2024-02-07Summary. Dear, Hello, according to the calculation you said, find that d is equal to 4 9, then the formula for the general term of an is equal to -1+(n-1)4 9
Dear, Hello, according to the calculation you said, find that d is equal to 4 9, then the formula for the general term of an is equal to -1+(n-1)4 9
Dear, Hello, according to the calculation you Bi Sell said Regret Hu teased d is equal to 4 9 Then the an general term formula is equal to -1 + (n-1)4 9 The sum formula is also a grind that can be directly substituted.
And what about the first n terms and sn?
sn is equal to n(a1+an) 2 You can directly substitute this equation Isn't the first n term sn.
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Anonymous users2024-02-06Summary. a1＋4d＝12
Get a5 12, a4 6, d 6
Let the sum of the first n terms of the difference series an be sn, a5=2a4, s9=108, and find the general term formula of the series an.
Let the sum of the first n terms of the equal difference series an be sn, a5=2a4, s9=108, and find the general formula for the series an: find a1 and d
a1+4d 2(a1+3d)9a1+36d108a1 4d 12 gives a5 12,a4 6,d 6a3 0 a2 -6 a1 -12
So, an -12 6n
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Anonymous users2024-02-05The formula for summing with the equal-yard balance series:
s4=4a1+4*3d 2,s2=a1+a2=2a1+d is obtained by s4=4s2.
4a1+4*3d 2=4*(2a1+d) is late.
d=2a1=2*1=2
The kernel orange is an=a1+(n-1)d=1+2*(n-1)=2n-1
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Anonymous users2024-02-04a4=s4-a3=16-5=11;
EquiHu difference: a4-a3=11-5=6
Then the pants are scattered, a1 = a3-2*6 = 5-12=-7 general term formula: reed.
an= a1+6n =-7+6n
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Anonymous users2024-02-03an is a series of equal differences, then a2n=a1+(2n-1)d; an=a1+（n-1）d
And because a2n=2an+1
So a2n=a1+(2n-1)d=2(a1+(n-1)d)+1, so there is a1+1=d
And because s4=4s2
So 2a1=d
There is a1=1;d=2
There is an=2n-1
1) a1+a12=a6+a7a1+a13=a7*2 can be written as the first term and the tolerance form can be used to prove that s12=(a1+a12)*12 2=(a6+a7)*6s13=(a1+a13)*13 2=a7*13, so a1+2d=12 a6+a7<0, that is, 2a1+11d>0 a7>0, that is, a1+6d<0 is represented by d by the formula A1, that is, a1=12-2d is brought into Eq. respectively: 24+7d>0 12+4d<0 can be solved to obtain -24 70a7<0 to know a6>0, a7<0, and |a6|>|a7|Therefore s1a7>a8>a12, so s6+a7>s7>s8>... >>>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
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
Since it is an equal difference series, so a8-a4=4d, d is the tolerance, then d=-4, from a4=a1+3d, we can know a1=a4-3d=24, from sn=na1+n(n-1)d 2 to get sn=-2n 2+26n >>>More
It is known that f(x)=a x+a x +a x +a n x , and a , a , a , a , .,a n is a series of equal differences, n is a positive and even number, and f(1)=n, f(-1)=n; Finding the general term of a n? >>>More