Miscellaneous Progressions
Some
Important Results:
- 1 + 2 + 3 +…+ n = (sum of
the first n natural numbers).
- 12 + 22
+ 32 + … + n2 = (sum of squares
of the first n natural numbers).
- 13+23+33+…+n3==(1+2+3+…+n)2
(sum of cubes of first n natural numbers).
- 1 + x + x2
+ x3 +........ = (1 - x)-1, if -1 < x <
1.
- 1 + 2x + 3x2
+........... = (1 - x)-2, if -1 < x < 1.
Method of Differences:
Suppose
a1, a2, a3, .......is a sequence such that the
sequence a2 - a1, a3 - a2,
.........is either an A.P. or a G.P. The n th term ‘an’ of this
sequence is obtained as follows :.
S = a1 +
a2 + a3 +........+an-1 + an.
S = a1
+ a2 +.........+an-2 + an-1 + an .
&⇒ an = a1 + [(a2
- a1) + (a3 - a2) +......+(an - an
- 1)].
Since
the terms within the brackets are either in an A.P. or in a G.P., we can find
the value of an, the nth term. We can now find the sum of the n
terms of the sequence
as
Example 1
Find the sum infinite
terms.
Solution:
Here tr =
= =
&⇒ sum of first n terms = =
&⇒
sum of infinite terms =
Example 2
Find the sum of Ist n terms of the
series 5, 7, 11, 17, 25,……..
Solution:
Let S = 5 +
7 + 11 + 17 + 25+……….+ tr.
S = 5 + 7 + 11
+ 17 + ….…… + tr – 1 + tr.
Subtracting, we
get 0 = 5 + 2 + 4 + 6 + 8 + ……..+ rth term – tr.
&⇒ tr = 5 + 2
tr = r2
– r + 5
Sn = = + 5n
= {(n + 1)(2n + 1) –
3(n + 1) + 30} = (n2 +
14).
·
To
find t1 + t2 + t3 + t4 + ….+ tn
.
Let Sn = t1 + t2
+ t3 + t4 + ….+ tn .
Let D
t1 , Dt2, Dt3, ….., Dtn–1 ( 1st
order difference ) .
D2 t1 , D2t2, D2t3, ….., D2tn–1 ( 2nd order difference ) .
…. …. …. .
…. …. …. .
Then tn = t1 + n–1C1D t1 + n–1C2D2t1 + …. +
n–1Cr–1Dn–1t1.
The result can be proved using mathematical
induction
Example 3
Find the nth term and sum to n terms
of the series 12, 40, 90, 168,
280, 432,…
Solution:
Given
series is
12, 40, 90, 168,
280, 432, ….
The given series
and the successive order difference are
t1 = 12,
40, 90, 168, 280, 432, ….
D t1 = 28, 50, 78, 112,
152, ….
D2 t1 = 22,
28, 34, 40, ….
D3 t1 = 6,
6, 6 ….
D4 t1 = 0,
0
Hence
nth terms, tn = 12 + 28 . n–1C1 + 22. n–1C2
+ 6.n–1C3 .
\ Sn =
= 12n
+ 28 nC2 + 22. nC3 + 6.nC4
.
= 12n
+ 28.
=
.