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Page 1
Ex.48 Sum to n terms of the series
) x 4 1 )( x 3 1 (
1
) x 3 1 )( x 2 1 (
1
) x 2 1 )( x 1 (
1
? ?
?
? ?
?
? ?
+......
Sol. Let T
r
be the general term of the series T
r
=
) x ) 1 r ( 1 )( rx 1 (
1
? ? ?
So T
r
= ?
?
?
?
?
?
? ? ?
? ? ? ?
) x ) 1 r ( 1 )( rx 1 (
) rx 1 ( ] x ) 1 r ( 1 [
x
1
= ?
?
?
?
?
?
? ?
?
? x ) 1 r ( 1
1
rx 1
1
x
1
T
r
= f(r) – f(r + 1)
? S = ? T
r
= T
1
+ T
2
+ T
3
+ ........ + T
n
= ?
?
?
?
?
?
? ?
?
? x ) 1 n ( 1
1
rx 1
1
x
1
=
] x ) 1 n ( 1 )[ x 1 (
n
? ? ?
Ex.49 (a) Sum the following series to infinity
7 · 4 · 1
1
+
10 · 7 · 4
1
+
13 · 10 · 7
1
+ ...........
(b) Sum the following series upto n-terms 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + ............
Sol. (a) T
n
=
] n 3 4 [ ] n 3 1 [ ] 3 ) 1 n ( 1 [
1
? ? ? ?
=
) 4 n 3 ) ( 1 n 2 ) ( 2 n 3 (
1
? ? ?
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
T
1
=
6
1
?
?
?
?
?
?
?
7 · 4
1
4 · 1
1
Page 2
Ex.48 Sum to n terms of the series
) x 4 1 )( x 3 1 (
1
) x 3 1 )( x 2 1 (
1
) x 2 1 )( x 1 (
1
? ?
?
? ?
?
? ?
+......
Sol. Let T
r
be the general term of the series T
r
=
) x ) 1 r ( 1 )( rx 1 (
1
? ? ?
So T
r
= ?
?
?
?
?
?
? ? ?
? ? ? ?
) x ) 1 r ( 1 )( rx 1 (
) rx 1 ( ] x ) 1 r ( 1 [
x
1
= ?
?
?
?
?
?
? ?
?
? x ) 1 r ( 1
1
rx 1
1
x
1
T
r
= f(r) – f(r + 1)
? S = ? T
r
= T
1
+ T
2
+ T
3
+ ........ + T
n
= ?
?
?
?
?
?
? ?
?
? x ) 1 n ( 1
1
rx 1
1
x
1
=
] x ) 1 n ( 1 )[ x 1 (
n
? ? ?
Ex.49 (a) Sum the following series to infinity
7 · 4 · 1
1
+
10 · 7 · 4
1
+
13 · 10 · 7
1
+ ...........
(b) Sum the following series upto n-terms 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + ............
Sol. (a) T
n
=
] n 3 4 [ ] n 3 1 [ ] 3 ) 1 n ( 1 [
1
? ? ? ?
=
) 4 n 3 ) ( 1 n 2 ) ( 2 n 3 (
1
? ? ?
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
T
1
=
6
1
?
?
?
?
?
?
?
7 · 4
1
4 · 1
1
T
2
=
6
1
?
?
?
?
?
?
?
10 · 7
1
7 · 4
1
? ? ?
T
n
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
——————————————————
? S
n
=
?
n
T =
6
1
?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 (
1
4 · 1
1
= ?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 ( 6
1
24
1
as n ? ? ? S
?
=
24
1
(b) 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + .............
T
n
=
5
1
n(n + 1)(n + 2)(n + 3) [(n + 4) – (n – 1)]
? T
1
=
5
1
[1 · 2 · 3 · 4 · 5 – 0]
T
2
=
5
1
[2 · 3 · 4 · 5 · 6 – 1 · 2 · 3 · 4 · 5]
? ? ?
T
n
=
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4) – (n – 1)n(n + 1)(n + 2)(n + 3)]
? S
n
=
?
n
T =
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4)] =
5
1
n(n + 1)(n + 2)(n + 3)(n + 4)
Ex.50 Find the sum to
n
terms and if possible also to infinity
1
1 3 7 . .
+
1
3 5 9 . .
+
1
5 7 1 1 . .
+ .......
Sol.
5
1 3 5 7 . . .
+
7
3 5 7 9 . . .
+
9
5 7 9 11 . . .
+ ....... ?
T
n
=
( )
( ) ( ) ( ) ( )
2 3
2 1 2 1 2 3 2 5
n
n n n n
?
? ? ? ?
=
( )
( ) ( ) ( ) ( )
2 5 2
2 1 2 1 2 3 2 5
n
n n n n
? ?
? ? ? ?
=
1
2 1 2 1 2 3 ( ) ( ) ( ) n n n ? ? ?
?
2
2 1 2 1 2 3 2 5 ( ) ( ) ( ) ( ) n n n n ? ? ? ?
=
1
4
1
2 1 2 1
1
2 1 2 3 ( ) ( ) ( ) ( ) n n n n ? ?
?
? ?
?
?
?
?
?
?
Page 3
Ex.48 Sum to n terms of the series
) x 4 1 )( x 3 1 (
1
) x 3 1 )( x 2 1 (
1
) x 2 1 )( x 1 (
1
? ?
?
? ?
?
? ?
+......
Sol. Let T
r
be the general term of the series T
r
=
) x ) 1 r ( 1 )( rx 1 (
1
? ? ?
So T
r
= ?
?
?
?
?
?
? ? ?
? ? ? ?
) x ) 1 r ( 1 )( rx 1 (
) rx 1 ( ] x ) 1 r ( 1 [
x
1
= ?
?
?
?
?
?
? ?
?
? x ) 1 r ( 1
1
rx 1
1
x
1
T
r
= f(r) – f(r + 1)
? S = ? T
r
= T
1
+ T
2
+ T
3
+ ........ + T
n
= ?
?
?
?
?
?
? ?
?
? x ) 1 n ( 1
1
rx 1
1
x
1
=
] x ) 1 n ( 1 )[ x 1 (
n
? ? ?
Ex.49 (a) Sum the following series to infinity
7 · 4 · 1
1
+
10 · 7 · 4
1
+
13 · 10 · 7
1
+ ...........
(b) Sum the following series upto n-terms 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + ............
Sol. (a) T
n
=
] n 3 4 [ ] n 3 1 [ ] 3 ) 1 n ( 1 [
1
? ? ? ?
=
) 4 n 3 ) ( 1 n 2 ) ( 2 n 3 (
1
? ? ?
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
T
1
=
6
1
?
?
?
?
?
?
?
7 · 4
1
4 · 1
1
T
2
=
6
1
?
?
?
?
?
?
?
10 · 7
1
7 · 4
1
? ? ?
T
n
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
——————————————————
? S
n
=
?
n
T =
6
1
?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 (
1
4 · 1
1
= ?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 ( 6
1
24
1
as n ? ? ? S
?
=
24
1
(b) 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + .............
T
n
=
5
1
n(n + 1)(n + 2)(n + 3) [(n + 4) – (n – 1)]
? T
1
=
5
1
[1 · 2 · 3 · 4 · 5 – 0]
T
2
=
5
1
[2 · 3 · 4 · 5 · 6 – 1 · 2 · 3 · 4 · 5]
? ? ?
T
n
=
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4) – (n – 1)n(n + 1)(n + 2)(n + 3)]
? S
n
=
?
n
T =
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4)] =
5
1
n(n + 1)(n + 2)(n + 3)(n + 4)
Ex.50 Find the sum to
n
terms and if possible also to infinity
1
1 3 7 . .
+
1
3 5 9 . .
+
1
5 7 1 1 . .
+ .......
Sol.
5
1 3 5 7 . . .
+
7
3 5 7 9 . . .
+
9
5 7 9 11 . . .
+ ....... ?
T
n
=
( )
( ) ( ) ( ) ( )
2 3
2 1 2 1 2 3 2 5
n
n n n n
?
? ? ? ?
=
( )
( ) ( ) ( ) ( )
2 5 2
2 1 2 1 2 3 2 5
n
n n n n
? ?
? ? ? ?
=
1
2 1 2 1 2 3 ( ) ( ) ( ) n n n ? ? ?
?
2
2 1 2 1 2 3 2 5 ( ) ( ) ( ) ( ) n n n n ? ? ? ?
=
1
4
1
2 1 2 1
1
2 1 2 3 ( ) ( ) ( ) ( ) n n n n ? ?
?
? ?
?
?
?
?
?
?
?
2
6
1
2 1 2 1 2 3
1
2 1 2 3 2 5 ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n ? ? ?
?
? ? ?
?
?
?
?
?
?
T
1
=
1
4
1
1 3
1
3 5 . .
?
?
?
?
?
?
?
?
1
3
1
1 3 5
1
3 5 7 . . . .
?
?
?
?
?
?
?
T
2
=
1
4
1
3 5
1
5 7 . .
?
?
?
?
?
?
?
?
1
3
1
3 5 7
1
5 7 9 . . . .
?
?
?
?
?
?
?
? S
n
=
1
4
1
1 3
1
2 1 2 3 . ( ) ( )
?
? ?
?
?
?
?
?
?
n n
?
1
3
1
1 3 5
1
2 1 2 3 2 5 . . ( ) ( ) ( )
?
? ? ?
?
?
?
?
?
?
n n n
S
?
=
1
1 3 .
1
4
1
15
?
?
?
?
?
?
?
=
11
6 0
.
1
3
=
11
1 8 0
and
S
n
=
1 1
1 8 0
?
1
2 1 2 3 ( ) ( ) n n ? ?
1
4
1
3 2 5
?
?
?
?
?
?
?
?
( ) n
=
1 1
1 8 0
?
6 1
12 2 1 2 3 2 5
n
n n n
?
? ? ? ( ) ( ) ( )
Ex.51 Evaluate the sum ,
1
1 2
1
k k k k r
k
n
( ) ( ) . . . . . . ( ) ? ? ?
?
?
.
Sol. S =
1
1 2 3 1
1
2 3 4 2 . . . . . . . ( ) . . . . . . . ( ) r r ?
?
?
+ ...... +
1
1 2 n n n n r ( ) ( ) . . . . . ( ) ? ? ?
T
1
=
1 1
1 2 3
1
2 3 4 1 r r r . . . . . . . . . . . . . . . . ( )
?
?
?
?
?
?
?
?
T
2
=
1 1
2 3 4 1
1
3 4 5 2 r r r . . . . . . . . ( ) . . . . . . . . ( ) ?
?
?
?
?
?
?
?
?
..
T
n
=
1 1
1 2 1
1
1 2 3 r n n n n r n n n n r ( ) ( ) . . . . . . ( ) ( ) ( ) ( ) . . . . . . ( ) ? ? ? ?
?
? ? ? ?
?
?
?
?
?
?
? Sum =
1 1
r r
n
n r !
!
( ) !
?
?
?
?
?
?
?
?
Page 4
Ex.48 Sum to n terms of the series
) x 4 1 )( x 3 1 (
1
) x 3 1 )( x 2 1 (
1
) x 2 1 )( x 1 (
1
? ?
?
? ?
?
? ?
+......
Sol. Let T
r
be the general term of the series T
r
=
) x ) 1 r ( 1 )( rx 1 (
1
? ? ?
So T
r
= ?
?
?
?
?
?
? ? ?
? ? ? ?
) x ) 1 r ( 1 )( rx 1 (
) rx 1 ( ] x ) 1 r ( 1 [
x
1
= ?
?
?
?
?
?
? ?
?
? x ) 1 r ( 1
1
rx 1
1
x
1
T
r
= f(r) – f(r + 1)
? S = ? T
r
= T
1
+ T
2
+ T
3
+ ........ + T
n
= ?
?
?
?
?
?
? ?
?
? x ) 1 n ( 1
1
rx 1
1
x
1
=
] x ) 1 n ( 1 )[ x 1 (
n
? ? ?
Ex.49 (a) Sum the following series to infinity
7 · 4 · 1
1
+
10 · 7 · 4
1
+
13 · 10 · 7
1
+ ...........
(b) Sum the following series upto n-terms 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + ............
Sol. (a) T
n
=
] n 3 4 [ ] n 3 1 [ ] 3 ) 1 n ( 1 [
1
? ? ? ?
=
) 4 n 3 ) ( 1 n 2 ) ( 2 n 3 (
1
? ? ?
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
T
1
=
6
1
?
?
?
?
?
?
?
7 · 4
1
4 · 1
1
T
2
=
6
1
?
?
?
?
?
?
?
10 · 7
1
7 · 4
1
? ? ?
T
n
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
——————————————————
? S
n
=
?
n
T =
6
1
?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 (
1
4 · 1
1
= ?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 ( 6
1
24
1
as n ? ? ? S
?
=
24
1
(b) 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + .............
T
n
=
5
1
n(n + 1)(n + 2)(n + 3) [(n + 4) – (n – 1)]
? T
1
=
5
1
[1 · 2 · 3 · 4 · 5 – 0]
T
2
=
5
1
[2 · 3 · 4 · 5 · 6 – 1 · 2 · 3 · 4 · 5]
? ? ?
T
n
=
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4) – (n – 1)n(n + 1)(n + 2)(n + 3)]
? S
n
=
?
n
T =
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4)] =
5
1
n(n + 1)(n + 2)(n + 3)(n + 4)
Ex.50 Find the sum to
n
terms and if possible also to infinity
1
1 3 7 . .
+
1
3 5 9 . .
+
1
5 7 1 1 . .
+ .......
Sol.
5
1 3 5 7 . . .
+
7
3 5 7 9 . . .
+
9
5 7 9 11 . . .
+ ....... ?
T
n
=
( )
( ) ( ) ( ) ( )
2 3
2 1 2 1 2 3 2 5
n
n n n n
?
? ? ? ?
=
( )
( ) ( ) ( ) ( )
2 5 2
2 1 2 1 2 3 2 5
n
n n n n
? ?
? ? ? ?
=
1
2 1 2 1 2 3 ( ) ( ) ( ) n n n ? ? ?
?
2
2 1 2 1 2 3 2 5 ( ) ( ) ( ) ( ) n n n n ? ? ? ?
=
1
4
1
2 1 2 1
1
2 1 2 3 ( ) ( ) ( ) ( ) n n n n ? ?
?
? ?
?
?
?
?
?
?
?
2
6
1
2 1 2 1 2 3
1
2 1 2 3 2 5 ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n ? ? ?
?
? ? ?
?
?
?
?
?
?
T
1
=
1
4
1
1 3
1
3 5 . .
?
?
?
?
?
?
?
?
1
3
1
1 3 5
1
3 5 7 . . . .
?
?
?
?
?
?
?
T
2
=
1
4
1
3 5
1
5 7 . .
?
?
?
?
?
?
?
?
1
3
1
3 5 7
1
5 7 9 . . . .
?
?
?
?
?
?
?
? S
n
=
1
4
1
1 3
1
2 1 2 3 . ( ) ( )
?
? ?
?
?
?
?
?
?
n n
?
1
3
1
1 3 5
1
2 1 2 3 2 5 . . ( ) ( ) ( )
?
? ? ?
?
?
?
?
?
?
n n n
S
?
=
1
1 3 .
1
4
1
15
?
?
?
?
?
?
?
=
11
6 0
.
1
3
=
11
1 8 0
and
S
n
=
1 1
1 8 0
?
1
2 1 2 3 ( ) ( ) n n ? ?
1
4
1
3 2 5
?
?
?
?
?
?
?
?
( ) n
=
1 1
1 8 0
?
6 1
12 2 1 2 3 2 5
n
n n n
?
? ? ? ( ) ( ) ( )
Ex.51 Evaluate the sum ,
1
1 2
1
k k k k r
k
n
( ) ( ) . . . . . . ( ) ? ? ?
?
?
.
Sol. S =
1
1 2 3 1
1
2 3 4 2 . . . . . . . ( ) . . . . . . . ( ) r r ?
?
?
+ ...... +
1
1 2 n n n n r ( ) ( ) . . . . . ( ) ? ? ?
T
1
=
1 1
1 2 3
1
2 3 4 1 r r r . . . . . . . . . . . . . . . . ( )
?
?
?
?
?
?
?
?
T
2
=
1 1
2 3 4 1
1
3 4 5 2 r r r . . . . . . . . ( ) . . . . . . . . ( ) ?
?
?
?
?
?
?
?
?
..
T
n
=
1 1
1 2 1
1
1 2 3 r n n n n r n n n n r ( ) ( ) . . . . . . ( ) ( ) ( ) ( ) . . . . . . ( ) ? ? ? ?
?
? ? ? ?
?
?
?
?
?
?
? Sum =
1 1
r r
n
n r !
!
( ) !
?
?
?
?
?
?
?
?
Ex.52 Sun to n terms of the series
5 . 4 . 3
6
4 . 3 . 2
5
3 . 2 . 1
4
? ?
+ ........
Sol. Let T
r
=
r 3
r(r 1)(r 2)
?
? ?
=
) 2 r )( 1 r ( r
3
) 2 r )( 1 r (
1
? ?
?
? ?
= ?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ) 2 r )( 1 r (
1
) 1 r ( r
1
2
3
2 r
1
1 r
1
? S= ?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
) 2 n )( 1 n (
1
2
1
2
3
2 n
1
2
1
= ?
?
?
?
?
?
?
?
?
?
) 1 n ( 2
3
1
2 n
1
4
5
=
) 2 n )( 1 n ( 2
1
4
5
? ?
?
[2n+ 5]
Ex.53 Sum of the series
3 . 2 . 1
n
+
4 . 3 . 2
1 n ?
+
5 . 4 . 3
2 n ?
+ .... +
) 2 n )( 1 n ( n
1
? ?
is
2
n 3
4(n 1)
?
?
Sol. t
r
, the rth term of the series is given by
t
r
=
) 2 r )( 1 r ( r
1 r n
? ?
? ?
= (n + 1)
) 2 r )( 1 r ( r
1
? ?
-
) 2 r )( 1 r (
1
? ?
= (n + 1)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2 r
1
1 r
1
) 2 r ( 2
1
1 r
1
r 2
1
[resolve into partial fractions]
= (n + 1) ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
2 r
1
1 r
1
2
1
1 r
1
r
1
2
1
– ?
?
?
?
?
?
?
?
? 2 r
1
1 r
1
=
2
1 n ?
?
?
?
?
?
?
?
?
1 r
1
r
1
–
?
?
?
?
?
?
?
?
1
2
1 n
?
?
?
?
?
?
?
?
? 2 r
1
1 r
1
? ?
?
n
1 r
r
t
=
?
?
?
?
?
? ?
2
1 n
?
?
?
?
?
?
?
?
1 n
1
1
–
?
?
?
?
?
? ?
2
3 n
?
?
?
?
?
?
?
?
2 n
1
2
1
=
2
n n 3 n 3 n 3
2 4 2(n 2) 4(n 2)
? ? ?
? ? ?
? ?
Ex.54 Find the sum of n terms of the series
... 2 .
5 . 4
3
2 .
4 . 3
2
2 .
3 . 2
1
3 2
? ? ?
Sol. Here u
n
=
n
2 .
) 2 n )( 1 n (
n
? ?
. Let
2 n
B
1 n
A
) 2 n )( 1 n (
n
?
?
?
?
? ?
, or, n = A(n + 2) + B(n + 1).
Equating the coeffs. of like powers of n, A + B = 1, 2A + B = 0. ? A = –1, B = 2.
Now we may write u
n
=
1 n
2
2 n
2
n 1 n
?
?
?
?
. Putting n = 1, 2, 3,...., n, we have e
u
1
=
2
2
3
2
2
? , u
2
=
3
2
4
2
2 3
? , u
3
=
4
2
5
2
3 4
? , ............... u
n
=
1 n
2
2 n
2
n 1 n
?
?
?
?
. By addition, S
n
=
2 n
2
1 n
?
?
– 1.
Page 5
Ex.48 Sum to n terms of the series
) x 4 1 )( x 3 1 (
1
) x 3 1 )( x 2 1 (
1
) x 2 1 )( x 1 (
1
? ?
?
? ?
?
? ?
+......
Sol. Let T
r
be the general term of the series T
r
=
) x ) 1 r ( 1 )( rx 1 (
1
? ? ?
So T
r
= ?
?
?
?
?
?
? ? ?
? ? ? ?
) x ) 1 r ( 1 )( rx 1 (
) rx 1 ( ] x ) 1 r ( 1 [
x
1
= ?
?
?
?
?
?
? ?
?
? x ) 1 r ( 1
1
rx 1
1
x
1
T
r
= f(r) – f(r + 1)
? S = ? T
r
= T
1
+ T
2
+ T
3
+ ........ + T
n
= ?
?
?
?
?
?
? ?
?
? x ) 1 n ( 1
1
rx 1
1
x
1
=
] x ) 1 n ( 1 )[ x 1 (
n
? ? ?
Ex.49 (a) Sum the following series to infinity
7 · 4 · 1
1
+
10 · 7 · 4
1
+
13 · 10 · 7
1
+ ...........
(b) Sum the following series upto n-terms 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + ............
Sol. (a) T
n
=
] n 3 4 [ ] n 3 1 [ ] 3 ) 1 n ( 1 [
1
? ? ? ?
=
) 4 n 3 ) ( 1 n 2 ) ( 2 n 3 (
1
? ? ?
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
T
1
=
6
1
?
?
?
?
?
?
?
7 · 4
1
4 · 1
1
T
2
=
6
1
?
?
?
?
?
?
?
10 · 7
1
7 · 4
1
? ? ?
T
n
=
6
1
?
?
?
?
?
?
? ?
?
? ? ) 4 n 3 ) ( 1 n 3 (
1
) 1 n 3 ) ( 2 n 3 (
1
——————————————————
? S
n
=
?
n
T =
6
1
?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 (
1
4 · 1
1
= ?
?
?
?
?
?
? ?
?
) 4 n 3 ) ( 1 n 3 ( 6
1
24
1
as n ? ? ? S
?
=
24
1
(b) 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + .............
T
n
=
5
1
n(n + 1)(n + 2)(n + 3) [(n + 4) – (n – 1)]
? T
1
=
5
1
[1 · 2 · 3 · 4 · 5 – 0]
T
2
=
5
1
[2 · 3 · 4 · 5 · 6 – 1 · 2 · 3 · 4 · 5]
? ? ?
T
n
=
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4) – (n – 1)n(n + 1)(n + 2)(n + 3)]
? S
n
=
?
n
T =
5
1
[n(n + 1)(n + 2)(n + 3)(n + 4)] =
5
1
n(n + 1)(n + 2)(n + 3)(n + 4)
Ex.50 Find the sum to
n
terms and if possible also to infinity
1
1 3 7 . .
+
1
3 5 9 . .
+
1
5 7 1 1 . .
+ .......
Sol.
5
1 3 5 7 . . .
+
7
3 5 7 9 . . .
+
9
5 7 9 11 . . .
+ ....... ?
T
n
=
( )
( ) ( ) ( ) ( )
2 3
2 1 2 1 2 3 2 5
n
n n n n
?
? ? ? ?
=
( )
( ) ( ) ( ) ( )
2 5 2
2 1 2 1 2 3 2 5
n
n n n n
? ?
? ? ? ?
=
1
2 1 2 1 2 3 ( ) ( ) ( ) n n n ? ? ?
?
2
2 1 2 1 2 3 2 5 ( ) ( ) ( ) ( ) n n n n ? ? ? ?
=
1
4
1
2 1 2 1
1
2 1 2 3 ( ) ( ) ( ) ( ) n n n n ? ?
?
? ?
?
?
?
?
?
?
?
2
6
1
2 1 2 1 2 3
1
2 1 2 3 2 5 ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n ? ? ?
?
? ? ?
?
?
?
?
?
?
T
1
=
1
4
1
1 3
1
3 5 . .
?
?
?
?
?
?
?
?
1
3
1
1 3 5
1
3 5 7 . . . .
?
?
?
?
?
?
?
T
2
=
1
4
1
3 5
1
5 7 . .
?
?
?
?
?
?
?
?
1
3
1
3 5 7
1
5 7 9 . . . .
?
?
?
?
?
?
?
? S
n
=
1
4
1
1 3
1
2 1 2 3 . ( ) ( )
?
? ?
?
?
?
?
?
?
n n
?
1
3
1
1 3 5
1
2 1 2 3 2 5 . . ( ) ( ) ( )
?
? ? ?
?
?
?
?
?
?
n n n
S
?
=
1
1 3 .
1
4
1
15
?
?
?
?
?
?
?
=
11
6 0
.
1
3
=
11
1 8 0
and
S
n
=
1 1
1 8 0
?
1
2 1 2 3 ( ) ( ) n n ? ?
1
4
1
3 2 5
?
?
?
?
?
?
?
?
( ) n
=
1 1
1 8 0
?
6 1
12 2 1 2 3 2 5
n
n n n
?
? ? ? ( ) ( ) ( )
Ex.51 Evaluate the sum ,
1
1 2
1
k k k k r
k
n
( ) ( ) . . . . . . ( ) ? ? ?
?
?
.
Sol. S =
1
1 2 3 1
1
2 3 4 2 . . . . . . . ( ) . . . . . . . ( ) r r ?
?
?
+ ...... +
1
1 2 n n n n r ( ) ( ) . . . . . ( ) ? ? ?
T
1
=
1 1
1 2 3
1
2 3 4 1 r r r . . . . . . . . . . . . . . . . ( )
?
?
?
?
?
?
?
?
T
2
=
1 1
2 3 4 1
1
3 4 5 2 r r r . . . . . . . . ( ) . . . . . . . . ( ) ?
?
?
?
?
?
?
?
?
..
T
n
=
1 1
1 2 1
1
1 2 3 r n n n n r n n n n r ( ) ( ) . . . . . . ( ) ( ) ( ) ( ) . . . . . . ( ) ? ? ? ?
?
? ? ? ?
?
?
?
?
?
?
? Sum =
1 1
r r
n
n r !
!
( ) !
?
?
?
?
?
?
?
?
Ex.52 Sun to n terms of the series
5 . 4 . 3
6
4 . 3 . 2
5
3 . 2 . 1
4
? ?
+ ........
Sol. Let T
r
=
r 3
r(r 1)(r 2)
?
? ?
=
) 2 r )( 1 r ( r
3
) 2 r )( 1 r (
1
? ?
?
? ?
= ?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ) 2 r )( 1 r (
1
) 1 r ( r
1
2
3
2 r
1
1 r
1
? S= ?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
) 2 n )( 1 n (
1
2
1
2
3
2 n
1
2
1
= ?
?
?
?
?
?
?
?
?
?
) 1 n ( 2
3
1
2 n
1
4
5
=
) 2 n )( 1 n ( 2
1
4
5
? ?
?
[2n+ 5]
Ex.53 Sum of the series
3 . 2 . 1
n
+
4 . 3 . 2
1 n ?
+
5 . 4 . 3
2 n ?
+ .... +
) 2 n )( 1 n ( n
1
? ?
is
2
n 3
4(n 1)
?
?
Sol. t
r
, the rth term of the series is given by
t
r
=
) 2 r )( 1 r ( r
1 r n
? ?
? ?
= (n + 1)
) 2 r )( 1 r ( r
1
? ?
-
) 2 r )( 1 r (
1
? ?
= (n + 1)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2 r
1
1 r
1
) 2 r ( 2
1
1 r
1
r 2
1
[resolve into partial fractions]
= (n + 1) ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
2 r
1
1 r
1
2
1
1 r
1
r
1
2
1
– ?
?
?
?
?
?
?
?
? 2 r
1
1 r
1
=
2
1 n ?
?
?
?
?
?
?
?
?
1 r
1
r
1
–
?
?
?
?
?
?
?
?
1
2
1 n
?
?
?
?
?
?
?
?
? 2 r
1
1 r
1
? ?
?
n
1 r
r
t
=
?
?
?
?
?
? ?
2
1 n
?
?
?
?
?
?
?
?
1 n
1
1
–
?
?
?
?
?
? ?
2
3 n
?
?
?
?
?
?
?
?
2 n
1
2
1
=
2
n n 3 n 3 n 3
2 4 2(n 2) 4(n 2)
? ? ?
? ? ?
? ?
Ex.54 Find the sum of n terms of the series
... 2 .
5 . 4
3
2 .
4 . 3
2
2 .
3 . 2
1
3 2
? ? ?
Sol. Here u
n
=
n
2 .
) 2 n )( 1 n (
n
? ?
. Let
2 n
B
1 n
A
) 2 n )( 1 n (
n
?
?
?
?
? ?
, or, n = A(n + 2) + B(n + 1).
Equating the coeffs. of like powers of n, A + B = 1, 2A + B = 0. ? A = –1, B = 2.
Now we may write u
n
=
1 n
2
2 n
2
n 1 n
?
?
?
?
. Putting n = 1, 2, 3,...., n, we have e
u
1
=
2
2
3
2
2
? , u
2
=
3
2
4
2
2 3
? , u
3
=
4
2
5
2
3 4
? , ............... u
n
=
1 n
2
2 n
2
n 1 n
?
?
?
?
. By addition, S
n
=
2 n
2
1 n
?
?
– 1.
Ex.55 Find the sum of the infinite series
.....
4 . 3
7
3 . 2
5
2 . 1
3
2 2 2 2 2 2
? ? ?
Sol. Here u
n
= 2 2
) 1 n ( n
1 n 2
?
?
=
2 2
2 2
) 1 n ( n
n ) 1 n (
?
? ?
=
2
n
1
– 2
) 1 n (
1
?
? ? ? u
1
=
2
1
1
–
2
2
1
, u
2
=
2
2
1
–
2
3
1
............... u
n
=
2
n
1
– 2
) 1 n (
1
?
By addition, S
n
= 1 – 2
) 1 n (
1
?
If n ? ??, then S
n
? 1. Hence the sum of the infinite series is 1.
Ex.56 Sum to n terms of the series
...
) 3 x )( 2 x )( 1 x (
x 3
) 2 x )( 1 x (
x 2
1 x
1
2
?
? ? ?
?
? ?
?
?
Sol. u
n
=
) n x )....( 2 x )( 1 x (
x x ) n x (
) n x )....( 2 x )( 1 x (
nx
n 1 n 1 n
? ? ?
? ?
?
? ? ?
? ?
=
) n x )....( 2 x )( 1 x (
x
) 1 n x )....( 2 x )( 1 x (
x
n 1 n
? ? ?
?
? ? ? ?
?
u
1
= 1 –
1 x
x
?
, u
2
=
) 2 x )( 1 x (
x
1 x
x
2
? ?
?
?
, u
3
=
) 3 x )( 2 x )( 1 x (
x
) 2 x )( 1 x (
x
3 2
? ? ?
?
? ?
.. .... .... .... .... ....
u
n
=
) n x )....( 2 x )( 1 x (
x
) 1 n x )...( 2 x )( 1 x (
x
n 1 n
? ? ?
?
? ? ? ?
?
? S
n
= 1 –
) n x )....( 2 x )( 1 x (
x
n
? ? ?
.
Ex.57 Let s
n
= 1 +
n
1
......
3
1
2
1
? ? ?
. Show that
(A) s
n
= n –
?
?
?
?
?
? ?
? ? ?
n
1 n
......
3
2
2
1
; (B) ns
n
= n +
?
?
?
?
?
?
?
?
?
? ?
?
?
?
1 n
1
2 n
2
.......
2
2 n
1
1 n
.
Sol. (A) It is obvious that
s
n
= 1 +
n
1
....
3
1
2
1
? ? ?
= n + ?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ? 1
n
1
....... 1
3
1
1
2
1
) 1 1 (
= n –
?
?
?
?
?
? ?
? ? ?
n
1 n
.....
3
2
2
1
.
(B) s
n
= ?
?
?
n k
1 k
k
1
, ns
n
= ? ?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
n k
1 k
n k
1 k
1
k
k n
k
k k n
. Hence, s
n
= n +
?
?
?
?
?
?
?
? ?
?
?
?
1 n
1
.....
2
2 n
1
1 n
.
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The content of Solved Examples for JEE: Summation of Series is prepared as per the latest JEE syllabus.
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