1.Algebra Booster

1.50 Algebra Booster
152. Let t
n
3n
+ 1
=
( n + 1)( n + 2)( n + 3)
3n
+ 3– 2
=
( n + 1)( n + 2)( n + 3)
3 2
= -
( n + 2)( n + 3) ( n + 1)( n + 2)(n + 3)
Ê 1 1 ˆ ( n + 3) -( n + 1)
= 3 Á - -
Ë ( n + 2) ( n + 3) ˜
¯ ( n + 1)( n + 2)( n + 3)
3 Ê 1 1 ˆ 1
= Á - -
Ë ( n + 2) ( n + 3) ˜
¯ ( n + 1)( n + 2)
1
+
( n + 2)( n + 3)
Ê 1 1 ˆ Ê 1 1 ˆ
= 3Á - - -
Ën + 2 n + 3˜ ¯
Á
Ën + 1 n + 2˜
¯
ÊÊ
1 1 ˆˆ
+ ÁÁ
-
Ën
+ 2 n + 3˜
¯
˜
Ë
¯
Ê 1 1 ˆ Ê 1 1 ˆ
= 4Á - - -
Ën + 2 n + 3˜ ¯
Á
Ën + 1 n + 2˜
¯
Thus,
ÊÊ1 1ˆ Ê1 1ˆ Ê 1 1 ˆˆ
S n
= 4 ÁÁ - ˜ + Á - ˜ +º+ -
3 4 4 5 n 2 n 3
˜
ËË ¯ Ë ¯ Á
Ë + + ˜
¯¯
ÊÊ1 1ˆ Ê1 1ˆ Ê 1 1 ˆˆ
– ÁÁ - ˜ + Á - ˜ + … + -
2 3 3 4 n 1 n 2
˜
ËË ¯ Ë ¯ Á
Ë + + ˜
¯¯
4 Ê1 1 ˆ Ê1 1 ˆ
= Á - - -
Ë 3 n + 3 ˜
¯
Á
Ë 2 n + 2 ˜
¯
Ê2
(3n
+ 5) ˆ
= Á -
Ë3 ( n + 2)( n + 3) ˜
¯
153. We have,
Sum =
1 1 1 1
+ + +º+
log 4 log 4 log 4 log 4
2 2 3
2 2 2 n
= log 4
2 + log 4
(2 2 ) + log 4
(2 3 ) + … + log 4
(2 n )
= log 4
(2 ◊ 2 2 ◊ 2 3 ◊ … ◊ 2 n )
= log 4
(2 1 + 2 + 3 + … + n )
1+ 2+ 3 + ... + n
= log 2 (2 )
2
Ê1+ 2+ 3+º+
nˆ
= Á log2
( 2
Ë
˜ )
2 ¯
Ê1+ 2+ 3 + … + nˆ
= Á
Ë
˜
2 ¯
nn ( + 1)
=
4
154. We have,
n
i
1
i= 1 j= 1k=
1
n i
j
ÂÂÂ
=
ÂÂ
i= 1 j=
1
n
i = 1
n
() j
Ênn
( + 1) ˆ
= ÂÁ Ë
˜
2 ¯
1 2
= Â( n + n)
2
i = 1
1 Ê
n n
2
ˆ
=
2
ÁÂn
+ Ân˜
Ëi= 1 i=
1 ¯
1 Ênn ( + 1) (2n+ 1) nn ( + 1) ˆ
= Á
+
2Ë
˜
6 2 ¯
1 nn ( + 1) Ê(2n+
1) ˆ
= ¥ Á + 1
2 2 Ë
˜
3 ¯
nn ( + 1)( n+
2)
=
6
155. Let S n
= 3 + 7 + 13 + 21 + … + T n – 1
+ T n
…(i)
Also, S n
= 3 + 7 + 13 + 21 + … + T n – 1
+ T n
…(ii)
Subtracting
(i) – (ii), we get
0 = 3 + 4 + + 6 + … + (T n
– T n – 1
) – T n
fi T n
= 3 + 4 + + 6 + … + (T n
– T n – 1
)
fi T n
= 3 + (4 + + 6 + … + (T n
– T n – 1
))
n - 1
= 3 + (2.4 + ( n -2)2)
2
= 3 + (n – 1)(n + 2)
= n 2 + n + 1
2
Therefore, S = ( n + n + 1)
n
Â
2
Ân
Ân
Â
= + + 1
nn ( + 1)(2n+ 1) nn ( + 1)
= + + n
6 2
2
nn ( + 3n+
5)
=
3
156. Given series is
1+ 3 + 7 + 15 + 31 + …
Here, the differences between the successive terms are
2, 4, 8, 16, …, which are in GP. Whenever the successive
differences are in GP, we consider its nth term as
t n
= ar n + br n–1 + c
In order to find the value of a, b and c, we put n = 1, 2, 3.
Thus, t 1
= 1 = a + b + c
t 2
= 3 = ar 2 + br + c,
and t 3
= 7 = ar 2 + br 2 + c