There are some series whose sum can be expressed by explicit formulae. Such series are called special series. 1. Sum of first n natural numbers 2. Sum of first n odd natural numbers 3. Sum of squares of first n natural numbers 4. Sum of cubes of first n natural numbers

**Special
Series**

There are some series
whose sum can be expressed by explicit formulae. Such series are called special series.

Here we study some
common special series like

(i) Sum of first â€˜*n*â€™
natural numbers

(ii) Sum of first â€˜*n*â€™
odd natural numbers.

(iii) Sum of squares of
first â€˜*n*â€™ natural numbers.

(iv) Sum of cubes of
first â€˜*n*â€™ natural numbers.

We can derive the
formula for sum of any powers of first *n* natural numbers using the
expression (*x* + 1)^{k}^{ +1} âˆ’ *x* ^{k}^{
+1} . That is to find 1* ^{k}* + 2

To find1 + 2 + 3 + + *n*
, let us consider the identity (*x* + 1)^{2} âˆ’ *x* ^{2}
= 2*x* + 1

When *x* = 1 , 2^{2}
âˆ’ 1^{2} = 2(1) + 1

When *x* = 2 , 3^{2}
âˆ’ 2^{2} = 2(2) + 1

When *x* = 3 , 4^{2}
âˆ’ 3^{2} = 2(3) + 1

** : :
:**

When *x* = *n*
âˆ’1 , *n* ^{2} âˆ’ (*n* âˆ’1)^{2} = 2(*n* âˆ’1) + 1

When *x* = *n*
âˆ’1 , (*n* + 1)^{2} - *n*
^{2}= 2(*n*) + 1

Adding all these
equations and cancelling the terms on the Left Hand side, we get,

(*n*
+ 1)^{2} âˆ’1^{2 }= 2(1 + 2 + 3 + + *n* ) + *n*

*n
*^{2}*
*+* *2*n*= 2(1 + 2 + 3 + + *n* ) + *n*

2(1 + 2 + 3 + + *n*) = *n* ^{2} + *n* = *n* (*n* + 1)

1 + 2 + 3 + + *n= *[*n *(*n
*+* *1)] / 2^{}

1 + 3 + 5 + + (2*n*
âˆ’1)

It is an A.P. with *a*
= 1 , *d* = 2 and *l* = 2*n* âˆ’1

To find1^{2} + 2^{2}
+ 3^{2} + + *n*^{2} , let us consider the identity (*x*
+ 1)^{3} âˆ’ *x* ^{3} = 3*x* ^{2} +
3*x* + 1

When *x* = 1 , 2 ^{3}
âˆ’ 1^{3} = 3(1)^{2} + 3(1) + 1

When *x* = 2 , 3 ^{3}
âˆ’ 2 ^{3} = 3(2)^{2} + 3(2) + 1

When *x* = 3 , 4 ^{3}
âˆ’ 3 ^{3} = 3(3)^{2} + 3(3) + 1

** : : :**

When *x* = *n*
âˆ’1 , *n* ^{3} âˆ’ (*n* âˆ’1)^{3} = 3(*n* âˆ’1)^{2}
+ 3(*n* âˆ’1) + 1

When *x* = *n*
, (*n* + 1)^{3} âˆ’*n*^{3} = 3*n* ^{2} + 3*n*
+ 1

Adding all these
equations and cancelling the terms on the Left Hand side, we get,

To find 1^{3} +
2^{3} + 3^{3} + + *n*^{3} , let us consider the
identity

(*x* + 1)^{4}
âˆ’ *x* ^{4} = 4*x* ^{3} + 6*x* ^{2}
+ 4*x* + 1

When *x* = 1 , 2 ^{4}
âˆ’ 1^{4 }= 4(1)^{2 }+ 4(1) + 1

When *x* =2 , 3^{4}
âˆ’ 2 ^{4 }= 4(2)^{3 }+ 4(2) + 1

When *x* = 3 , 4^{4}
âˆ’ 3 ^{4 }= 4(3)^{3} + 6(3) + 1

: : :

When *x* = *n*
âˆ’1 ,*n*^{4} âˆ’ (*n* âˆ’1)^{4 }= 4(*n* âˆ’1)^{3}
+ 4(*n* âˆ’1) + 1

When *x* = *n*
, (*n* + 1)^{4} âˆ’*n*^{4 }= 4*n*^{3}
+ 6*n*^{2} + 4*n* + 1

Adding all these
equations and cancelling the terms on the Left Hand side, we get,

(*n*+1)^{4}â€“1^{4}
= 4(1^{3} + 2^{3 }+ 3 ^{3} + + *n* ^{3}
) + 6(1^{2} + 2^{2} + 3^{2} + + *n*^{2}
) + 4(1 + 2 + 3 + + *n* ) + *n*

*n*^{4}* *+* *4*n*^{3}*
*+* *6*n*^{2}* *+* *4*n *=* *4(1^{3}
+ 2^{3} + 3 ^{3} + + *n* ^{3} ) + 6 Ã—

4(1^{3} + 2^{3}
+ 3^{3} + â€¦ + *n*^{3} ) = *n*^{4
}+ 4*n*^{3} + 6*n* ^{2} +4*n* âˆ’ 2*n*^{3}
âˆ’*n*^{2} âˆ’ 2*n*^{2} âˆ’ *n* âˆ’ 2*n*^{2}
âˆ’ 2*n* â€“*n*

4(1^{3} + 2^{3}
+ 3^{3} + â€¦ + *n*^{3} ) = *n*^{4
}+ 2*n*^{3} + *n* ^{2} = *n*^{2} (*n*^{2}
+ 2*n +*1) = *n ^{2}*(n+1)

Consider the numbers
220 and 284.

Sum of the divisors of
220 (excluding 220) = 1+2+4+5+10+11+20+22+44+55+110=284

Sum of the divisors of
284 (excluding 284) =1+2+4+71+142=220.

Thus, sum of divisors
of one number excluding itself is the other. Such pair of numbers is called **Amicable Numbers **or **Friendly Numbers.**

220 and 284 are least
pair of Amicable Numbers. They were discovered by Pythagoras. We now know more
than 12 million amicable pair of Numbers.

**Think Corner: 1. **How many squares are
there in a standard chess board? 2. How many rectangles are there in a standard
chess board?

Here is a summary of
list of some useful summation formulas which we discussed.

These formulas are used
in solving summation problems with finite terms.

Find the value of (i) 1** **+** **2** **+** **3** **+** **...** **+** **50** **(ii)** **16** **+** **17** **+** **18** **+** **...** **+** **75

(i) 1+** **2

Using, 1 + 2 + 3 +
+ *n = *

1+ 2 + 3 + + 50 =

(ii) 16 + 17 + 18 + +
75 = (1 + 2 + 3 + + 75) âˆ’(1 + 2 +
3 + + 15)

=75(75 + 1)/2 _{âˆ’}
15(15 + 1) / 2

=2850 âˆ’120 = 2730

Find the sum of

(i) 1 + 3 + 5 + â€¦ +
to 40 terms

(ii) 2 + 4 + 6 + â€¦
+ 80

(iii) 1+3 + 5 + â€¦ +
55

(i) 1+3** **+

(ii) 2 + 4 + 6 + â€¦ + 80
= 2(1 + 2 + 3 + â€¦ + 40) = 2 Ã— [40 Ã— (40 + 1)]/2 = 1640

(iii) 1 + 3 + 5 + â€¦
+ 55

Here the number of terms
is not given. Now we have to find the number of terms using the
formula, *n* = (*l-a*)/*d* + 1 gives *n= *[(55-1)/2] + 1 = 28

Therefore, 1 + 3 + 5
+ + 55 = (28)^{2} = 784

Find the sum
of

(i) 1^{2} + 2^{2}
+ + 19^{2}

(ii) 5^{2 }+ 1 0^{2
}+ 15^{2}_{ }+ + 105^{2 }

(iii) 15^{2} +
16^{2 }+ 17 ^{2} + + 28^{2}

Find the sum of (i) 1^{3}
+ 2^{3} + 3^{3} + + 16^{3} (ii) 9^{3} +
10^{3} + + 21^{3}* *

**Example 2.58** If** **1** **+** **2** **+** **3** **+ â€¦ +** ***n*** **=** **666** **then find** ***n*.

Since,** **1

*n *^{2}* *+* n *âˆ’1332* *=*
*0* *gives* *(*n *+* *37 )(*n *âˆ’* *36)*
*=* *0

So, *n* = âˆ’37 or *n*
= 36

But *n* â‰ âˆ’37
(Since *n* is a natural number);

Hence *n* = 36.

**Progress Check**

Say
True or False. Justify them.

Â·
The sum of first *n* odd natural numbers
is always an odd number.

Â·
The sum of consecutive even numbers is always
an even number.

Â·
The difference between the sum of squares of
first *n* natural numbers and the sum of first *n* natural numbers is
always divisible by 2.

Â·
The sum of cubes of the first *n* natural
numbers is always a square number.

Tags : Definition, Theorem, Example, Solution | Mathematics , 10th Mathematics : UNIT 2 : Numbers and Sequences

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10th Mathematics : UNIT 2 : Numbers and Sequences : Special Series | Definition, Theorem, Example, Solution | Mathematics

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