Number series problems math masters number series problems. A master who loves mathematics.

I ask you, can you add parentheses, t(n-1)*t(n+1)=t(n)*t(n)+5

Is that the case?

(1)s[n]=(a[1]+a[n])n/2=(a[1]+a[1]+(n-1)d)n/2=(a[1]+4d+a[1]+(n-5)d)n/2

a[5]+a[n-4])n 2=16n, so n=136 16=, the question is wrong?

If you answer 21, then s[n]=16*21=336

2) take n=11,s(a[11])=(a[1]+a[11])n 2=a[6]*n=38,s(b[n])=(b[1]+b[11])n 2=b[6]*n=19

s(a[11])/s(b[11])=a[6]/b[6]=38/19=2

3) Because (n+1) 3=n 3+3n 2+3n+1

So (n+1) 3-n 3=3n 2+3n+1

So n 3-(n-1) 3=3(n-1) 2+3(n-1)+1

So 2 3-1 3=3*1 2+3*1+1

It all adds up to:

n+1)^3-1=3(1^2+2^2+..n^2)+3(1+2+..n)+n

So 3 (1 2 + 2 2 +.n^2)=(n+1)^3-3n(n+1)/2-(n+1)=(n+1)[2(n+1)^2-3n-2]/2

n+1)[2n^2+n]/2=n(n+1)(2n+1)/2

So 1 2 + 2 2 + 3 2 + 4 2 +...n^2=n(n+1)(2n+1)/6

Upstairs, you have question 1 n=9??So how do you explain a[5]=2,a[9-4]=a[5]=30?

I asked and asked. = Down.

1. sn=(a1+an)n 2=136 and a5+a(n-4)=a1+an=32, i.e. 16n=136 get n

The title is wrong, how can I calculate that n is not an integer?

There is no problem with the question, it is a problem of strength upstairs, and it is not explained.

1.Let the first term be A1 and the tolerance is D, according to the stem information, there are A1+4D=2, A1+(N-5)D=30, Na1+N(N-1)D 2=136. The simultaneous three-formula eliminates a1 and d, and the simplification has 2n 2-35n + 153 = 0

solution n=9 or rounded), so n=9

2.Let the first terms of the series be a1 and b1 respectively, and the tolerances are d1 and d2 respectively according to the title of [na1+n(n-1)d1 2] [nb1+n(n-1)d2 2]=(3n+5) (2n-3), and simplify that is, [nd1+(2a1-d1)] [nd2+(2b1-d2)]=(3n+5) (2n-3).

The corresponding identity has d1=3, d2=2, a1=4, b1=-1 2, so a6 b6=(a1+5d1) (b1+5d2)=(4+5*3) (-1 2+5*2)=2

n n*) Although you can't be superstitious about authority, you must first ensure your own accuracy.

This problem requires the first term of the sequence to be solved. Let's set the first item to A0.

an=2a(n-1)+3(n-1)

4a(n-2)+2*3*(n-2)+3*(n-1)

2^n*a0+3*[1*2^(n-2)+2*2^(n-3)+.n-1)*2^0]

is the s in the middle parentheses, then.

2s=1*2^(n-1)+2*2^(n-2)+.n-1)*2^1

s= 1*2^(n-2)+.n-2)*2^1+(n-1)*2^0

Subtract item by item with the above alignment to get.

s=2s-s=2^(n-1)+2^(n-2)+.2^1-(n-1)=2^n-2-n+1=2^n-1-n

So an=2 n*(a0+3)-3-3*n

A series that is both an equal difference series and an equal proportional series must be a non-zero constant series. So the common ratio is 1(Tolerance is 0).

Is it really possible? Let the three numbers a+d a+2d and calculate according to the law of proportional sequences....

Ask a question, point r(n-1 n, 1 n).

1）.Does n-1 n express n-(1 n) or (n-1) n?

2）.Here n is 1, 2, 3, 4 ......Or is it the same as n in "real number m,n"?

You make it clear, I'll see if I can do it

a(n+1)+a(n+2)+.a(2n)=100-30=70a1+a2+..an=30

Subtract and get. n*d*n=40

And the first 3n terms and for.

The first 2n terms and +a(2n+1)+a(2n+2)+a(3n) and a(2n+1)+a(2n+2)+a(3n)=(a1+2n*d)+(a2+2n+d)+.

an+2n*d)=(a1+a2+..an)+2n*d*n=30+80=110, so the sum of the first 3n terms is 110+100=210