Posted on - 14-02-2017

IIT JEE

Some Important Results:

- 1 + 2 + 3 +…+ n = (sum of the first n natural numbers).
- 1
^{2}+ 2^{2}+ 3^{2}+ … + n^{2}= (sum of squares of the first n natural numbers). - 1
^{3}+2^{3}+3^{3}+…+n^{3}==(1+2+3+…+n)^{2}(sum of cubes of first n natural numbers). - 1 + x + x
^{2}+ x^{3}+........ = (1 - x)^{-1}, if -1 < x < 1. - 1 + 2x + 3x
^{2}+........... = (1 - x)^{-2}, if -1 < x < 1.

Suppose
a_{1}, a_{2}, a_{3}, .......is a sequence such that the
sequence a_{2} - a_{1}, a_{3} - a_{2},
.........is either an A.P. or a G.P. The n th term ‘a_{n}’ of this
sequence is obtained as follows :.

S = a_{1} +
a_{2} + a_{3} +........+a_{n-1} + a_{n}.

S = a_{1}
+ a_{2} +.........+a_{n-2} + a_{n-1} + a_{n }.

&⇒ a_{n} = a_{1} + [(a_{2}
- a_{1}) + (a_{3} - a_{2}) +......+(a_{n} - a_{n}
_{- 1})].

Since
the terms within the brackets are either in an A.P. or in a G.P., we can find
the value of a_{n}, the nth term. We can now find the sum of the n
terms of the sequence

as

Find the sum infinite terms.

Here t_{r} =

= =

&⇒ sum of first n terms = =

&⇒
sum of infinite terms =

Find the sum of Ist n terms of the series 5, 7, 11, 17, 25,……..

Let S = 5 +
7 + 11 + 17 + 25+……….+ t_{r}.

S = 5 + 7 + 11
+ 17 + ….…… + t_{r – 1} + t_{r}.

Subtracting, we
get 0 = 5 + 2 + 4 + 6 + 8 + ……..+ rth term – t_{r}.

&⇒ t_{r} = 5 + 2

t_{r} = r^{2}
– r + 5

S_{n} = = + 5n

= {(n + 1)(2n + 1) –
3(n + 1) + 30} = (n^{2} +
14).

·
To
find t_{1} + t_{2} + t_{3} + t_{4} + ….+ t_{n}
.

Let S_{n} = t_{1} + t_{2}
+ t_{3} + t_{4} + ….+ t_{n} .

Let D
t_{1} , Dt_{2}, Dt_{3}, ….., Dt_{n–1} ( 1^{st}
order difference ) .

D^{2} t_{1} , D^{2}t_{2}, D^{2}t_{3}, ….., D^{2}t_{n–1} ( 2^{nd} order difference ) .

…. …. …. .

…. …. …. .

Then t_{n} = t_{1} + ^{n–1}C_{1}D t_{1} + ^{n–1}C_{2}D^{2}t_{1} + …. +
^{n–1}C_{r–1}D^{n–1}t_{1}.

The result can be proved using mathematical induction

Find the nth term and sum to n terms
of the series 12, 40, 90, 168,

280, 432,…

Given series is

12, 40, 90, 168, 280, 432, ….

The given series and the successive order difference are

t_{1} = 12,
40, 90, 168, 280, 432, ….

D t_{1} = 28, 50, 78, 112,
152, ….

D^{2} t_{1} = 22,
28, 34, 40, ….

D^{3} t_{1} = 6,
6, 6 ….

D^{4} t_{1} = 0,
0

Hence
nth terms, t_{n} = 12 + 28 . ^{n–1}C_{1} + 22. ^{n–1}C_{2}
+ 6.^{n–1}C_{3} .

\ S_{n} =

= 12n
+ 28 ^{n}C_{2} + 22. ^{n}C_{3} + 6.^{n}C_{4}
.

= 12n + 28.

= .