Everything Maths Grade 12

CHAPTER 3. SEQUENCES AND SERIES 3.7

3. The common difference of an arithmetic series is 3. Calculate the value of n for which the nth
   term of the series is 93 , and the sum of the first n terms is 975.


4. The sum of n terms of an arithmeticseries is 5 n^2 − 11 n for all values of n. Determine the
   common difference.


5. The third term of anarithmetic sequence is− 7 and the 7 thterm is 9. Determine the sum ofthe
   first 51 terms of the sequence.


6. Calculate the sum ofthe arithmetic series 4 + 7 + 10 +··· + 901.

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(1.) 01cj (2.) 01ck (3.) 01cm (4.) 01cn (5.) 01cp (6.) 01cq

3.7 Finite Squared Series EMCZ

When we sum a finite number of terms in a quadratic sequence, we geta finite quadratic series. The

general form of a quadratic series is quite complicated, so we will onlylook at the simple casewhen

D = 2 and d = (a 2 −a 1 ) = 3, where D is the common seconddifference and d is the first difference.

This is the sequence of squares of the integers:

ai = i^2

{ai} ={ 12 ; 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, we write

Sn=

�n

i=1

i^2 = 1 + 4 + 9 +··· +n^2

which can be written, ingeneral, as

Sn=

�n

i=1

i^2 =

n

6

(2n + 1)(n + 1) (3.26)

The proof for Equation (3.26) can be found under the Extension block that follows:

Extension: Derivation of the FiniteSquared Series

We will now prove the formula for the finite squared series:

Sn=

�n

i=1

i^2 = 1 + 4 + 9 +··· +n^2

We start off with the expansion of (k + 1)^3.

(k + 1)^3 = k^3 + 3k^2 + 3k + 1

(k + 1)^3 −k^3 = 3k^2 + 3k + 1