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How to determine the formula of the sum of a series given its $n$th-term formula like: $$U_n= n^2+n$$

or

$$U_n= 6n^2 -12n + 5$$

QC_QAOA
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David HM
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2 Answers2

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It is known that $\sum_{k=1}^nk=1+2+3+...+n=\frac{n(n+1)}{2}$, and $\sum_{k=1}^n{k^2}=1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

By using these fact, we can evaluate the sum you are talking about.

Let $U_n=n^2+n$

$S_n=\sum_{k=1}^nU_k=\sum_{k=1}^n{k^2+k}=\sum_{k=1}^nk^2+\sum_{k=1}^nk$

$S_n=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$

Simplify and we get

$S_n=\frac{n(n+1)(n+2)}{3}$

Furthermore, if you want to calculate the sum of cube of first $n$ natural number,

$1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2=\frac{n^2(n+1)^2}{4}$

The main idea is to separate each term based on their power $(2,2^2,$ or $2^3)$ and calculate each sum with the formula.

For $U_n=\frac1{n^2+n}=\frac{n+1-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac1{n}-\frac1{n+1}$

$S_n=U_1+U_2+U_3+...+U_n=(\frac11-\frac12)+(\frac12-\frac13)+(\frac13-\frac14)+...+(\frac1n+\frac1{n+1})$

$\frac12$ and $-\frac12$ will cancel each other out. The same thing happens for $\frac13,\frac14,\frac15,...,\frac1n$. Thus the remaining terms are $S_n=\frac11-\frac1{n+1}=\frac{(n+1)-1}{n+1}=\frac{n}{n+1}$

sentheta
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  • Thanks. But how if the terms are a denominator like in Un= 1/(n^2+n), how to separate each of them? – David HM Jan 26 '20 at 04:45
  • For Un=1/(n^2+n), I usually use factorizing and separating the term so that telescoping is able to be done. Un=1/n(n+1)=(1/n)-(1/(n+1)). By summing up their value, lots of term are going to cancel out and leave us with (usually) simple arithmetic. – sentheta Jan 26 '20 at 06:00
  • I tried but stuck. Could you please do it? – David HM Jan 26 '20 at 08:11
  • Un=1/(n(n+1))=((n+1)-n)/(n(n+1)=(n+1)/(n(n+1))-n/(n(n+1))=(1/n)-(1/(n+1)). Then evaluate Sn. Sn=U1+U2+U3+...+Un=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/(n+1)). Consider the fact that there are -1/2 and +1/2 that will cancel each other our. The same thing happens for 1/3, 1/4, 1/5, ..., 1/n. Thus the remaining terms are Sn=1/1-1/(n+1)=n/(n+1) – sentheta Jan 26 '20 at 09:02
  • Thanks. I din't use that +n-n, that's why I was stuck. I usually use that technique in mathematical induction. Didn't realize that the technique is also useful in sequence and series topic – David HM Jan 26 '20 at 14:55
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First you need the forward difference operator on a function f $$ \Delta f(x)=f(x+1)-f(x)$$.So if you are given a function g and you need to sum it from a t0 b, you find f such that $g(x)=\Delta f(x)$. Then $$ \sum_{i=a}^b g(i)=\sum_{i=a}^b \Delta f(i)= f(b+1)-f(a) $$ Then you need falling factorials $n^{(i)}=n(n-1)...(n-i+1) $ , the relations $ \Delta n^{(i)}=in^{(i-1)} $ and $ n^{(i)}=\Delta(((1/(i+1))n^{(i+1)}) $.Then you need to convert a polynomial to a polynomial in falling factorials. Use $ n^{(1)}=n^1=n$ and $ n^{(i)}n=n^{(i+1)}+in^{(i)}$ so $n^2=n^{(2)}+n^{(1)}$ and $$ n^3=n^2n=(n^{(2)}+n^{(1)})n=n^{(2)}n+n^{(1)}n =n^{(3)}+2n^{(2)}+n^{(2)}+n^{(1)}$$ $$ =n^{(3)}+3n^{(2)}+n^{(1)}$$ So $ \sum_{x=a}^b x^3 =\sum_{(x=a)}^b x^{(3)}+3x^{(2)}+x^{(1)}$ $$ =\sum_{x=a}^b \Delta ((1/4)x^{(4)}+x^{(3)}+(1/2)x^{(2)}) $$ $$=(1/4)(b+1)b(b-1)(b-2)+(b+1)b(b-1)+(1/2)(b+1)b$$ $$-((1/4)a(a-1)(a-2)(a-3)+a(a-1)(a-2)+(1/2)a(a-1)) $$

P. Lawrence
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