# What is the summation of the series 1,3,6,10,15,….?

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Alright, looking at this series, see how the amount by which each term increases is one more than the previous term’s amount?

Another way to say it is that the
increments are in arithmetic progression. This is also called a second level AP.

Once that little tidbit became clear, it was easy enough to come up with a way to express the nth term by adding the sum of the progression to the initial value, in this case, 1.

Once the nth term can be expressed in terms of n, I simply took the sum of it by known identities (i.e. sum of r, r^2 etc to for n terms…)

In the given sequence, the differences of consecutive terms

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$\text{In the given sequence, the differences of consecutive terms}$

follow an Arithmetic progression

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$\text{follow an Arithmetic progression}$

Let

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$\text{Let}$

S=1+3+6+10+15++Tn1+Tn

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$S=1+3+6+10+15+\cdots +{T}_{n-1}+{T}_{n}$

S=       1+3+6+10+  +Tn2+Tn1+Tn

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Subtract the two equations

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$\text{Subtract the two equations}$

0=1+2+3+4+5++nTn

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$0=1+2+3+4+5+\cdots +n-{T}_{n}$

Tn=1+2+3+4n=n(n+1)2

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$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{T}_{n}=1+2+3+4\cdots n=\frac{n\left(n+1\right)}{2}$

Sn=k=1nTk=k=1nk(k+1)2=16k=1n(k+1)(k)((k+2)(k1))

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$\therefore {S}_{n}=\sum _{k=1}^{n}{T}_{k}=\sum _{k=1}^{n}\frac{k\left(k+1\right)}{2}=\frac{1}{6}\sum _{k=1}^{n}\left(k+1\right)\left(k\right)\left(\left(k+2\right)-\left(k-1\right)\right)$

Sn=16(k=1n(k)(k+1)(k+2)(k1)(k)(k+1))

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$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{S}_{n}=\frac{1}{6}\left(\sum _{k=1}^{n}\left(k\right)\left(k+1\right)\left(k+2\right)-\left(k-1\right)\left(k\right)\left(k+1\right)\right)$

Sn=16(1.2.30.1.2+2.3.41.2.3n(n+1)(n+2)(n1)n(n+1))

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$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{S}_{n}=\frac{1}{6}\left(1.2.3-0.1.2+2.3.4-1.2.3\cdots n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\right)$

Sn=16(n(n+1)(n+2))

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$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{S}_{n}=\frac{1}{6}\left(n\left(n+1\right)\left(n+2\right)\right)$

1+3+6+10+15n terms=n(n+1)(n+2)6

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$\overline{)\overline{)\therefore 1+3+6+10+15\cdots \text{n terms}=\frac{n\left(n+1\right)\left(n+2\right)}{6}}}$

Question : What is the summation of the series 1,3,6,10,15,….?

Answer : The series is relatively easy. We see that every number in the series is the sum of the preceding number in the series and its successive number in the integer scale.

I.e. 1, 3, 6, 10, 15 … is nothing but 1, 1+2, 1+2 =3+3, 3+3 = 6 + 4, 6 + 4 = 10 +5 …

In the series of 1, 3, 6, 10, 15…, the succeeding numbers are the sums of the preceding numbers with 2 then 3 then 4 then 5. In this vein, we can also add 6, then 7 then 8 then 9 then 10 then 11 to the preceding number and deduce the sequence as, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66 and so on and so forth. One can easily deduce the pattern in this sequence by deducting each predecessor from its successor.

1+3+6+10+15++n2(n+1)=t=1nk2(k+1)

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$\phantom{\rule{1em}{0ex}}1+3+6+10+15+\cdots +\frac{n}{2}\cdot \left(n+1\right)\phantom{\rule{0ex}{0ex}}=\sum _{t=1}^{n}\frac{k}{2}\left(k+1\right)$

=12(k=1nk2+k=1nk)=12[n6(n+1)(2n+1)+n2(n+1)]

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$=\frac{1}{2}\left(\sum _{k=1}^{n}{k}^{2}+\sum _{k=1}^{n}k\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\frac{n}{6}\left(n+1\right)\left(2n+1\right)+\frac{n}{2}\left(n+1\right)\right]$

=n12(n+1)(2n+1+3)=n6(n+1)(n+2)

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$=\frac{n}{12}\left(n+1\right)\left(2n+1+3\right)\phantom{\rule{0ex}{0ex}}=\frac{n}{6}\left(n+1\right)\left(n+2\right)$

21

1+2=3

3+3=6

6+4=10

10+5=15

15+6=21

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If you mean sum of infinite terms then this series is divergent and it will be infinity

Trivial answers are using Arithmetical Progression(A.P.) and Natural Number sum formula. Lets be a little creative now:

let the sum be S:

then,

S = 2 + 3 + 4 + …….+ 99 + 100 ( 99 terms)

S = 100 + 99 + 98 + …… + 3 + 2 (99 terms)

——————————————

2S = 102 + 102 + 102………. + 102 +102 (99 times)

2S = 102*99

S = 102*99 / 2

S = 5049

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These are triangle numbers. The sum of the listed sequence is 35, but to make this more exciting, let’s say you want to find a general expression for the sum of the n first triangle numbers!

The expression for the nth triangle number can be found using polynomial interpolation, yielding the formula:

an=12n2+12n=n(n+1)2

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${a}_{n}=\frac{1}{2}{n}^{2}+\frac{1}{2}n=\frac{n\left(n+1\right)}{2}$

Now, one can define:

s1=1

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${s}_{1}=1$

s2=1+3=4

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${s}_{2}=1+3=4$

s3=1+3+6=10

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${s}_{3}=1+3+6=10$

s4=1+3+6+10=20

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${s}_{4}=1+3+6+10=20$

and so on…

Using polynomial interpolation on these 4 points will yield a polynomial p(x)

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$p\left(x\right)$

of degree 3 such that p(n)=sn

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$p\left(n\right)={s}_{n}$

. This yields the following polynomial:

sn=p(n)=16n3+12n2+13n=n(n+1)(n+2)6

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${s}_{n}=p\left(n\right)=\frac{1}{6}{n}^{3}+\frac{1}{2}{n}^{2}+\frac{1}{3}n=\frac{n\left(n+1\right)\left(n+2\right)}{6}$

So there you have it. The formula for the sum of the n first triangle numbers is

sn=n(n+1)(n+2)6

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${s}_{n}=\frac{n\left(n+1\right)\left(n+2\right)}{6}$

Some interesting things to note:

1. Any finite sequence can be expressed as a polynomial of finite degree.

2. Let q(x)=k=1xp(k)

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$q\left(x\right)=\sum _{k=1}^{x}p\left(k\right)$

, and let p(x)

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$p\left(x\right)$

be a polynomial. Then q(x)

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$q\left(x\right)$

is a polynomial such that degq(x)=degp(x)+1

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$\mathrm{deg}q\left(x\right)=\mathrm{deg}p\left(x\right)+1$

. This means that the sum of any polynomial can be written in closed form.

3. When polynomial interpolation is used, at least n+1

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$n+1$

points are needed to find one and only one polynomial of degree n that passes through all the points.

The sum of the terms of a sequence is called a series .

If a sequence is arithmetic or geometric here are formulas to find the sum of the first n terms, denoted Sn, without actually adding all of the terms.