Consider the sequence 1, 4, 9, 16, 25, ….

a) Write down an expression forS._{n}

b) FindSfor_{n}n=1, 2, 3, 4, and 5.

✍ Solution:

a)S=1_{n}^{2}+2^{2}+3^{2}+4^{2}+…+n^{2}{all terms are squares}

b)

S_{1}=1

S_{2}=1+4=5

S_{3}=1+4+9=14

S_{4}=1+4+9+16=30

S_{5}=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) 1

^{2}+2

^{2}+3

^{2}+…+

*n*

^{2}(sum of squares of the first

*n*natural numbers)

(iii) 1

^{3}+2

^{3}+3

^{3}+…+

*n*

^{3}(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*=½

_{n}*n*(

*n*+1). (💎 See post • Calculating the sum of each Arithmetic Series from its sigma notation 👈)

(ii) Here

*S*=1

_{n}^{2}+2

^{2}+3

^{2}+…+

*n*

^{2}

We consider the identity (

*k*-1)

^{3}=

*k*

^{3}-3

*k*

^{2}+3

*k*-1

*k*

^{3}-(

*k*-1)

^{3}=3

*k*

^{2}-3

*k*+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}*i*

^{2}

{

*a*}={1

_{i}^{2}, 2

^{2}, 3

^{2}, 4

^{2}, 5

^{2}, 6

^{2}, …}

={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}

*i*

^{2}=1+4+9+⋯+

*n*

^{2}

which can be written, in general, as

*S*=∑

_{n}^{n}

_{i=1}

*i*

^{2}=⅙(2

*n*+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}

*i*

^{2}=1+4+9+⋯+

*n*

^{2}.

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}2

*j*

^{2}

✍ Solution:

The formula of choice is ∑

^{n}

_{i=1}

*i*

^{2}=⅙

*n*(

*n*+1)(2

*n*+1). In this application, we see

^{25}

_{j=1}

*j*

^{2}=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} (2*k*^{2}+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}(2

*k*

^{2}+5)=∑

^{10}

_{k=1}2

*k*

^{2}+∑

^{10}

_{k=1}5

=2∑

^{10}

_{k=1}

*k*

^{2}+∑

^{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}

*k*

^{2}=⅙⋅10⋅11⋅21=385. Similarly ∑

^{10}

_{k=1}5=10⋅5=50.

Putting all that together, ∑

^{10}

_{k=1}(2

*k*

^{2}+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)=

*n*

^{2}+3

*n*.

Thus, the sum to

*n*terms is given by

*S*=∑

_{n}^{n}

_{k=1}

*a*=∑

_{k}^{n}

_{k=1}(

*n*

^{2}+3

*n*)=∑

^{n}

_{k=1}

*k*

^{2}+3∑

^{n}

_{k=1}k

=⅙

*n*(2

*n*+1)(

*n*+1)+3⋅½

*n*(

*n*+1)

=⅓

*n*(

*n*+1)(

*n*+5)

📌 Ex4. Find the sum to *n* terms of the series: 5+11+19+29+41+⋯

✍ Solution: Let us write

*S*=5+11+19+29+…+

_{n}*a*

_{(n-1)}+

*a*

_{n}or

*S*= 5+11+19+…+

_{n}*a*

_{(n-2)}+

*a*

_{(n-1)}+

*a*.

_{n}On subtraction, we get

*n*-1) terms]-

*a*

_{n}Remember the sum of an AP

S=½n[2_{n}a_{1}+d(n-1)]. HenceS_{(n-1)}=½(n-1)[2a_{1}+d(n-2)].

*a*=5+½(

_{n}*n*-1)[2⋅6+2(

*n*-2)]

*a*=5+(

_{n}*n*-1)(

*n*+4)

*a*=

_{n}*n*

^{2}+3

*n*+1.

Hence

*S*=∑

_{n}^{n}

_{k=1}

*a*=∑

_{k}^{n}

_{k=1}(

*k*

^{2}+3

*k*+1)=∑

^{n}

_{k=1}

*k*

^{2}+3∑

^{n}

_{k=1}

*k*+

*n*

*n*(2

*n*+1)(

*n*+1)+3⋅½

*n*(

*n*+1)+

*n*

=⅓

*n*(

*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 “2*n*”, you would respond with the listing 2, 4, 6, 8 … and if I asked for entries in a sequence with general term “*n*^{2}”, 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:

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,

1^{3}+2^{3}+3^{3}+…+*n*^{3} (sum of cubes of the first *n* natural numbers).

💎 Let us read post •Cubic Series👈.