It is known that the formula for the first N terms and how to find it

For "The formula for the general term of a known series of numbers is an=(2 n-1) 2 n, where the sum of the first n terms is 321 64 Find n? The question can be done like this:

an=(2 n-1) 2 n=1-(1 2) n, then.

sum(an)=n-sum((1 2) n), followed by the sum of proportional sequences, you can use the formula directly, then.

sum(an)=n-(1 2)*[1-(1 2) n] [1-1 2]=n-1+(1 2) n, observe the above equation, and find that the result is a fraction with a denominator of 2 n, and the numerator is (n-1)*2 n+1 is an odd number, irreducible, then the comparison result is 321 64, you can get:

2 n=64, then n=6

For "If the formula for the general term of the series is an=2 n+2n-1, what is the sum of the first n terms of the series?" ”

It can be calculated directly, divided into three sums, the first term is a proportional series, the second result is 2n*n, and the third term is -n

Then add all three and you're done

an=1-1/2^n

sn=a1+a2+a3+……an=1-1/2+1-1/4+1-1/8……+1-1/2^n=n-(1/2+1/4+1/8+……1/2^n)

Let bn=1 2 n, which is a proportional series.

The common ratio is 1 2, tn=b1(1-(1 2) n) (1-(1 2)=1-(1 2) n

So sn=n-(1-(1 2) n)=n+(1 2) n-1=321 64

n=6an=2 n+2n-1, which can be seen as a proportional series and an equal difference series.

The sum of the same. sn=(2+4+8+……2^n)+(1+3+5+……2n-1)=2(1-2^n)/(1-2)+n(a1+an)/2=2^(n+1)-2+n^2

Since sn = 2n -3, there is also a laugh

sn-1 = 2(n-1)² 3

So Liang Qinmu:

an = sn - sn-1

2[n² -n-1)²]

2(2n - 1)

4n - 2

a1 = s1 = 2 * 1² -3 = 1 ≠ 4 * 1 - 2 = 2

Therefore, the general term of Oak Sen AN is:

a1 = 1, n = 1

an = 4n - 2 , n ≥2

Summary. Hello, it is a pleasure to answer for you - - the general term formula of the equal difference series.

an=a1＋(n－1)d

Promotional. an=am+(n－m)d

The first n terms of the equal difference series and the formula.

sn＝（a1+an）*n/2

sn=na1+n(n-1)d/2

Proportional sequence of general term formulas.

General formula: an=a1*q (n 1);

Promotional: an=am·q (n m);

Summation formula: sn=na1(q=1).

sn=[a1(1-q)^n]/(1-q)

Hope mine can help you, hope].

Know how to find the sum of the first n terms in the general term formula of a series.

Hello, it is a pleasure to answer for you - - the general term formula of the equal difference series.

an=a1＋(n－1)d

Promotional. an=am+(n－m)d

The first n terms of the equal difference series and the formula.

sn＝（a1+an）*n/2

sn=na1+n(n-1)d/2

Proportional sequence of general term formulas.

General formula: an=a1*q (n 1);

Promotional: an=am·q (n m);

Summation formula: sn=na1(q=1).

sn=[a1(1-q)^n]/(1-q)

Hope mine can help you, hope].

I won't use it. Wait a minute.

Classmate's formula tells you that you can't yet.

I don't know how to apply it to the problem, the formula has basically been memorized, I just don't know how to expand, should I write out the first n terms and each of the terms, or should I use another method?

Write out the first n items and write.

How to find the sum of the first n terms: Use the reverse order addition method to find the sum of the first n terms of the sequence.

For example, if the sum of the two items equidistant from the first and last terms is equal to the sum of the first and last terms, you can add the two sums of the two sums written upright and backwards to get the sum of a constant sequence, and this summing method is called inverted addition. When we learn knowledge, we should not only know its effect, but also find its cause, the process of knowledge is the source of knowledge, and it is also a tool for studying the same type of knowledge, such as the series of equal differences.

The derivation of the first n terms and formulas is "reverse addition".

Example 1: Let the difference series, the tolerance is d, and verify the first n terms of sn=n(a1+an) 2 Analysis:

sn=a1+a2+a3+..an ①

In reverse order: sn=an+an-1+an-2+....+a1 ②

Get: 2sn=(a1+an)+(a2+an-1)+(a3+an-2)+....an+a1) and a1+an=a2+an-1=a3+an-2=....=an+a1∴2sn=n(a2+an) sn=n(a1+an)/2

Dial: From the derivation process, it can be seen that the reason for the application of the reverse addition method is with the help of a1+an=a2+an-1=a3+an-2=....=an+a1, i.e., the sum of the two terms equidistant from the first and last terms is equal to the sum of the first and last terms.

Equations for the difference series.

Equations for the difference series.

The formula for the difference series is an=a1+(n-1)d

The sum of the first n terms is: sn=na1+n(n-1)d 2 if the tolerance d=1: sn=(a1+an)n 2 if m+n=p+q: am+an=ap+aq, if m+n=2p, then: am+an=2ap

The above n are positive integers.

Text translation. The value of the nth term an = first term + (number of terms - 1) tolerance.

The sum of the first n terms sn=first term + last term Number of terms (number of terms-1) tolerance 2 tolerance d=(an-a1) (n-1).

Number of Items = (Last Item - First Term) Tolerance + 1

When the number column is an odd number, the sum of the first n terms = the number of intermediate terms.

The number column is an even number of terms, find the first and last terms, add the first and last terms, divide it by the sum of 2 equal differences, and the formula for the middle term is 2an+1=an+an+2, where is the equal difference series.

There are two methods, one, the general term is multiplied by the proportional coefficient, and two, the series method, and the answer is 3

An is equal to the sum of the first n terms and minus the first n-1 terms, i.e., an sn s(n-1).

The first n + 1 terms of the number series and the sum of the first n terms of the number series.