Multiply 2n by n times of 3 to find the formula for the sum of the nth terms

Summary. Equal or proportional.

The sum of the first n terms of n times n times of 3n times -2.

Equal or proportional.

Proportional. Good.

I changed the title to the picture.

That's right. Isn't this an in your figure equal to each other, is this your original question?

Yes, it should be summed with groupings.

That's got it here.

Summary. The sum of the first n terms is 2n (n-1) of sn=3. The sum of the first n terms is sn=3n(n-1).

The derivation process is as follows: s(2n-1)=a1*(2n-1)+[2n-1)*(2n-2)*3] 2=3*(2n-1)+(2n-4n-2n+2)*3 2=6n-3+3n+9n+3=3n-3n=3n(n-1).

The sum of the first n terms is sn=3n(n-1). The derivation process is as follows: s(2n-1)=a1*(2n-1)+[2n-1)*(2n-2)*3] 2=3*(2n-1)+(2n-4n-2n+2)*3 2=6n-3+3n+9n+3=3n=3n=3n(n-1).

The sum of the first n terms is sn=3n(n-1).

is the sum of n terms of 2n- to the power of 2 - 1 of sn=3.

It's the same kiss.

Not sn 3n (n 1).

The first n term and Sun Pei's laughing group are 2n (n-1) of sn=3. The sum of the first n terms is sn=3n(n-1). The process of derivation is as follows:

s(2n-1)=a1*(2n-1)+[2n-1)*(2n-2)*3]/2=3*(2n-1)+(2n²-4n-2n+2)*3/2=6n-3+3n²-9n+3=3n²-3n=3n(n-1)

The detailed operation process is shown in the number of grandchildren

For reference, please smile.

This kind of solution belongs to the category of 'will do, and it is difficult to destroy the first to do it right.' Be careful.

Make use of misplaced subtraction.

s=1×1/3+3×1/3^2+5×1/3^3+…+2n-1) 1 3 n, multiply 1 3 on both sides at the same time, and get a new formula, 1 3s=....

The two formulas subtract the revelation, sort out the liquid quietly, and then find the s.

an=3 n+2n+1, let bn=3 n (proportional series) b1=3q=3cn=2n+1 (equal difference bucket rock column) c1=3d=2 The sum of the first n terms of the series: an=bn+cn=3*(3 n-1) (3-1)+(1 suspicion 2)*(3+2n+1)n=(3 2)*(3 n-1)+(n+2)n = (celery mold 1 2)*3 (n+1)-(3 2)+n 2+2n=(1 2)*3 (n+1)+n 2+2n-(3). 2) (n...

The formula of the Lingzu Ziqi Wangtong Spike Cluster of the nth power of 3 is the general term formula step of the nth power-2n: an=sn-s(n-1)=3n 2-2n-3(n-1) 2+2(n-1)=3n 2-2n-3n 2+6n-3+2n-2=6n-2.

Use the sum and disadvantages to call the annihilation phase subtraction method: sn=1*3 1+3*3 2+5*3 3++2n-1)*3^n3*sn= 1*3^2+3*3^3+.

2n-3)*3^n+(2n-1)*3^(n+1)-2sn=1*3^1+ 2*3^2+2*3^3+.+2*3^n-(2n-1)*3^(n+1)=3+2*(3^2+3^3+.+3^n)-(2n-1)*3^(n+1)=3-9+3^(n+1)-(2n-1)*3^(n+1)..