Suppose, If you have a big series of numbers, and someone asked you the sum of the series, then what will you do? Will you add every number of the series one by one? I think this is not an effective and efficient way.
So, for solving these types of problems, we use SUMMATION.
Let’s take the formula of summation and try to understand it.
Sol. i=110i=1+2+…+10
=1+2+3+4+5+6+7+8+9+10
Sol. i=132n-1=2*1-1 +2*2-1+(2*3-1)
= 9
These are the simple examples if you have a typical problem then,
You can solve it by two methods:-
Method-1
Simply you put it in summation formula and add expression terms
i=15i2 + i + 1
But this a very lengthy method
Method-2
We also have some results generated by using Arithmetic and Geometric Progressions.
It is a type of sequence, in which the difference between two consecutive terms is constant.
This is known as the common difference of AP.
Example,
3,9,12,15,18,… is an AP.
Its first term is 2 which is denoted by a.
The common difference is 3, denoted by d and it is the same for every two consecutive terms, that’s why it is an AP.
It is a type of sequence, in which consecutive terms have a constant ratio. This constant ratio is called Common Ratio.
Example,
5,25,125,… Is a GP
a = 2
common ratio denoted by r is 2. It is the same for every consecutive term, that’s why it is a GP.
Are you wondering, why we are talking about AP & GP in Summation?
Well, I want to inform you that, AP and GP has their formula for doing the sum of sequences. So, these two may help us in finding a summation of typical series.
S = n22a+n-1d or S = n2a+l (sum of all terms of AP)
a = First term of Sequence
n = no of terms in the sequence
d = common difference of AP (d = a2 – a1 )
l = last term of the sequence
Sn = arn-1r-1 ( r ≠ 1, r > 1) or Sn = a1-rn(1-r) (r ≠ 1 , r <1)
a = First term of GP
n = no of terms in GP
r = common ratio of GP (r = a2 / a1 )
If GP series is inifinite, s=a1-r
1.Find p=13p2
Sol. Method 1:-
p=1np2= nn+12n+16
= 33+16+16
= 14
Method 2:-
p=13p2= 12 + 22 +32
= 14
Let’s do some tricky questions based on summations.
Sol. i=12(j=13(xj+ai))
= i=12(j=13xj+j=13ai )
= i=12(x1+x2+x3+3ai)
= i=12(0+2+1+3ai)
=i=12(3+3ai)
=i=12(3)+3i=12( ai)
=3(2) + 3 (a1+ a2+ a3)
=6 + 3 (1 + 0 + (-1) )
= 6 + 0
= 6
i=11259
Sol. i=11259=9+9+9+…+9
(This Number (9) will be add 125 times, so there is a simple method for doing this and that is multiplication, which we have also done in the above problem .)
. i=11259=9×125
= 1125
16
p=1np+1p+5= 16n(n+7)(2n+7)
Sol. L.H.S = p=1np+1p+5
= p=1np2+6p+5
= p=1np2+p=1n6p+p=1n5
= 16n(n+1)(2n+1) + 6 × 12 n (n+1) + 5n
= 16n [ n+12n+1+18n+1+30]
= 16n [2n2+3n+1+18n+18+30]
= 16n[2n2+21n+49]
=16n(n+7)(2n+7)
Hence Proved.
Summation is a very useful technique to sum the larger expressions or series efficiently. It will also reduce the mistakes which can happen in calculating large series. We have also discussed Arithmetic and geometric progressions, which can be helpful in dealing with some type of series. We have also discussed the sum of Arithmetic progression and geometric progression and explained the summation with the working examples. The definition of the index of summation and the history related to summation has already been discussed. By knowing about summation, you will be able, sum up, typical sequences and series as well.