Consider the sequence 1, 4, 9, 16, 25, ….
a) Write down an expression for S_n.
b) Find S_n for n=1, 2, 3, 4, and 5.
✍ Solution:
a) S_n=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) S_n=1+2+3+…+n, then S_n=½n(n+1). (💎 See post • Calculating the sum of each Arithmetic Series from its sigma notation 👈)
(ii) Here S_n=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

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

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:

a_i=i²
{a_i }={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
S_n=∑ⁿ_i=1 i²=1+4+9+⋯+n²
which can be written, in general, as
S_n=∑ⁿ_i=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:

S_n=∑ⁿ_i=1 i²=1+4+9+⋯+n².
We start off with the expansion of (k+1)³.

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

📌 Example 1. Evaluate ∑²⁵_j=1 2j²
✍ Solution:
The formula of choice is ∑ⁿ_i=1 i²=⅙n(n+1)(2n+1). In this application, we see
2∑²⁵_j=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 ∑¹⁰_k=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.

∑¹⁰_k=1 (2k²+5)=∑¹⁰_k=1 2k²+∑¹⁰_k=1 5
=2∑¹⁰_k=1 k² +∑¹⁰_k=1 5.
The two sums We have left, can be found using formulas 1 and 3 above!
We see that ∑¹⁰_k=1 k²=⅙⋅10⋅11⋅21=385. Similarly ∑¹⁰_k=1 5=10⋅5=50.
Putting all that together, ∑¹⁰_k=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 a_n=n(n+3)=n²+3n.
Thus, the sum to n terms is given by

S_n=∑ⁿ_k=1 a_k=∑ⁿ_k=1 (n²+3n)=∑ⁿ_k=1 k² +3∑ⁿ_k=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

S_n=5+11+19+29+…+a_(n-1)+a_n
or S_n= 5+11+19+…+a_(n-2)+a_(n-1)+a_n.
On subtraction, we get
0=5+[6+8+10+12+⋯ (n-1) terms]-a_n

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

a_n=5+½(n-1)[2⋅6+2(n-2)]
a_n=5+(n-1)(n+4)
a_n=n²+3n+1.
Hence S_n=∑ⁿ_k=1 a_k=∑ⁿ_k=1 (k²+3k+1)=∑ⁿ_k=1 k² +3∑ⁿ_k=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👈.