How to Calculate a Quadratic Series within Sigma Notation?

Consider the sequence 1, 4, 9, 16, 25, ….
a) Write down an expression for Sn.
b) Find Sn for n=1, 2, 3, 4, and 5.
✍ Solution:
a) Sn=12+22+32+42+…+n2 {all terms are squares}
b)

S1=1
S2=1+4=5
S3=1+4+9=14
S4=1+4+9+16=30
S5=1+4+9+16+25=55

Sum to n Terms of Special Series
We shall now find the sum of first n terms of some special series, namely;
(i) 1+2+3+…+n (sum of first n natural numbers)
(ii) 12+22+32+…+n2 (sum of squares of the first n natural numbers)
(iii) 13+23+33+…+n3 (sum of cubes of the first n natural numbers).
Let us take them one by one.

(i) Sn=1+2+3+…+n, then Snn(n+1). (💎 See post 👉 Calculating the sum of each Arithmetic Series from its sigma notation 👈)
(ii) Here Sn=12+22+32+…+n2
We consider the identity (k-1)3=k3-3k2+3k-1

k3-(k-1)3=3k2-3k+1

Putting k=1, 2, …, n successively, we obtain
quadratic series a

If we add all the terms on the right and left, we arrive at
quadratic series b

Finite Squared Series
When we sum a finite number of terms in a quadratic sequence, we get a finite quadratic series. The general form of a quadratic series is quite complicated, so we will only look at the simple case below. This is the sequence of squares of the integers:

ai=i2
{ai }={12, 22, 32, 42, 52, 62, …}
={1, 4, 9, 16, 25, 36, …}

If we wish to sum this sequence and create a series, then we write
Sn=∑ni=1 i2=1+4+9+⋯+n2

which can be written, in general, as
Sn=∑ni=1 i2=⅙(2n+1)(n+1) _ (iv)

The proof for equation (iv) can be found under the Advanced block that follows:

Derivation of the Finite Squared Series
We will now prove the formula for the finite squared series:

Sn=∑ni=1 i2=1+4+9+⋯+n2.

We start off with the expansion of (k+1)3.
quadratic series c

If we add all the terms on the right and left, we arrive at
quadratic series d

📌 Example 1. Evaluate ∑25j=1 2j2
✍ Solution:
The formula of choice is ∑ni=1 i2=⅙n(n+1)(2n+1). In this application, we see
2∑25j=1 j2=2⋅⅙⋅25(25+1)(2⋅25+1)
=⅓⋅25⋅26⋅51
=25⋅26⋅17
=11050.

(Note the use of the basic math technique “cancellation.”)

📌 Ex2. Find the value of the sum ∑10k=1 (2k2+5) .

First, we’ll use the properties above to split this into two sums, then factor the 2 out of the first sum.

10k=1 (2k2+5)=∑10k=1 2k2+∑10k=1 5
=2∑10k=1 k2 +∑10k=1 5.

The two sums We have left, can be found using formulas 1 and 3 above!
We see that ∑10k=1 k2=⅙⋅10⋅11⋅21=385. Similarly ∑10k=1 5=10⋅5=50.
Putting all that together, ∑10k=1 (2k2+5)=2⋅385+50=820.

📌 Ex3. Find the sum to n terms of the series whose nth term is n(n+3).
✍ Solution: Given that an=n(n+3)=n2+3n.
Thus, the sum to n terms is given by

Sn=∑nk=1 ak=∑nk=1 (n2+3n)=∑nk=1 k2 +3∑nk=1 k
=⅙n(2n+1)(n+1)+3⋅½n(n+1)
=⅓n(n+1)(n+5)

📌 Ex4. Find the sum to n terms of the series: 5+11+19+29+41+⋯
✍ Solution: Let us write

Sn=5+11+19+29+…+a(n-1)+an
or Sn= 5+11+19+…+a(n-2)+a(n-1)+an.

On subtraction, we get
0=5+[6+8+10+12+⋯ (n-1) terms]-an

Remember the sum of an AP Sn=½n[2a1+d(n-1)]. Hence S(n-1)=½(n-1)[2a1+d(n-2)].

an=5+½(n-1)[2⋅6+2(n-2)]
an=5+(n-1)(n+4)
an=n2+3n+1.

Hence Sn=∑nk=1 ak=∑nk=1 (k2+3k+1)=∑nk=1 k2 +3∑nk=1 k+ n
=⅙n(2n+1)(n+1)+3⋅½n(n+1)+n
=⅓n(n+2)(n+4).

📌 Ex5. We are often given a “general term” or “nth term” for a sequence. For example, if I told you that you to list entries in a sequence with general term “2n”, you would respond with the listing 2, 4, 6, 8 … and if I asked for entries in a sequence with general term “n2”, you would respond with 1, 4, 9, 16, 25, … . What about the sum of the first 10 entries of such a sequence? Use the formula (iv) above:

⅙⋅10⋅(10+1)⋅(20+1)=⅙⋅10⋅11⋅21

You would then evaluate using the arithmetic tools of cancellation and calculation to obtain 385.

(iii). We shall find the sum of first n terms of a cubic series,
13+23+33+…+n3 (sum of cubes of the first n natural numbers).
💎 Let us read post 👉Cubic Series👈.

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