Sequences are important question types. By practicing a few questions one can easily crack these questions with ease. Study the following.
1. Find the sum to n terms of series 1.2.3 + 2.3.4 + 3.4.5+ ........
a. (n+1)(n+2)(n+3)/3
b. n(n+1)(2n+2)(n+2)/4
c. n(n+1)(n+2)
d. n(n+1)(n+2)(n+3)/4
Sol:
The general term of the above series = n(n+1)(n+2)
Sum = $\sum {n(n + 1)(n + 2) = \sum {{n^3} + 3{n^2} + 2n = \sum {{n^3} + 3\sum {{n^2}} } } } + 2\sum n $
( Recap the formulas for the sum of first n natural numbers, and its squares and its cubes)
$ \Rightarrow \displaystyle\frac{{{n^2}{{(n + 1)}^2}}}{4} + 3.\displaystyle\frac{{n(n + 1)(2n + 1)}}{6} + 2.\displaystyle\frac{{n(n + 1)}}{2}$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{{n^2} + n + 4n + 2 + 4}}{4}} \right]$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{n(n + 1)}}{4} + \frac{{2n + 1}}{2} + 1} \right]$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{{n^2} + 5n + 6}}{4}} \right]$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{(n + 2)(n + 3)}}{4}} \right]$
So option 4.
Alternate method:
Above mentioned method to be used only when the sum upto a certain terms was asked. Say upto 20 terms. If infinite terms sum was asked, Simply we check the answer options. Take the sum upto 3 terms.
6 + 24 + 60 = 90.
Now substitute n = 3 in the answer options. Only Option 4 satisfies.
2. $\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} + \sqrt {1 + \displaystyle\frac{1}{{{2^2}}} + \displaystyle\frac{1}{{{3^2}}}} + ..........\sqrt {1 + \displaystyle\frac{1}{{{{2007}^2}}} + \displaystyle\frac{1}{{{{2008}^2}}}} $ =
a. 2008 $ - \displaystyle\frac{1}{{2008}}$
b. 2007 $ - \displaystyle\frac{1}{{2007}}$
c. 2007 $ - \displaystyle\frac{1}{{2008}}$
d. 2008 $ - \displaystyle\frac{1}{{2007}}$
Sol:
The clue to solve the problem lies in the options provided. Observe that there are total 2007 terms given. and options are in 2007's and 2008's
Consider only first term:
$\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} $ = $\sqrt {1 + 1 + \displaystyle\frac{1}{4}} = \sqrt {\displaystyle\frac{9}{4}} = \displaystyle\frac{3}{2} = 2 - \displaystyle\frac{1}{2}$
Consider first two terms:
$\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} + \sqrt {1 + \displaystyle\frac{1}{{{2^2}}} +\displaystyle \frac{1}{{{3^2}}}} $
$ \Rightarrow $ $\displaystyle\frac{3}{2} + \sqrt {1 + \frac{1}{4} + \frac{1}{9}} = \frac{3}{2} + \sqrt {\frac{{49}}{{36}}} = \frac{3}{2} + \frac{7}{6} = \frac{8}{3} = 3 - \frac{1}{3}$
Similarly: $\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} + \sqrt {1 + \displaystyle\frac{1}{{{2^2}}} + \displaystyle\frac{1}{{{3^2}}}} + ..........\sqrt {1 + \displaystyle\frac{1}{{{{2007}^2}}} + \displaystyle\frac{1}{{{{2008}^2}}}} $ = 2008 - $\displaystyle\frac{1}{{2008}}$
3. S= $\displaystyle\frac{1}{{1 + {1^2}}} + \frac{1}{{2 + {2^2}}} + \frac{1}{{3 + {3^2}}} + ........$ What is the value of S?
(1) 2
(2) 1.5
(3) 1
(4) The sum is not finite.
Sol: Give S= $\displaystyle\frac{1}{{1 + {1^2}}} + \frac{1}{{2 + {2^2}}} + \frac{1}{{3 + {3^2}}} + ........$
$ \Rightarrow \displaystyle\frac{1}{{1(1 + 1)}} + \frac{1}{{2(1 + 2)}} + \frac{1}{{3(1 + 3)}} + ........$
$ \Rightarrow \displaystyle\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ........$
$ \Rightarrow \left( {\displaystyle\frac{1}{1} - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ........$
$ \Rightarrow \displaystyle\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + .......$
4. N, the set of natural numbers, is partitioned into subsets ${S_{_1}}$ = {1}, ${S_{_2}}$ = {2, 3}, ${S_{_3}}$={4, 5, 6}, ${S_{_4}}$ = {7, 8, 9, 10} ...... Find the sum of the elements in the subset ${S_{50}}$
Sol: Observe that the last term of every subset is 1, 1 +2, 1+2+3, 1+2+3+4, ....
So last term is written as $\displaystyle\frac{{n(n + 1)}}{2}$
Now nth subset has n elements in Arithmetic Progression with common difference 1.
To get the first term we have to substract (n-1) from the last term. (take S4, if we substract 3 from 7, we get the first term of that subset)
So first term = $\displaystyle\frac{{n(n + 1)}}{2} - (n - 1) = \displaystyle\frac{{{n^2} + n - 2n + 2}}{2} = \displaystyle\frac{{{n^2} - n + 2}}{2}$
So sum = $\displaystyle\frac{n}{2}\left[ {\displaystyle\frac{{{n^2} - n + 2}}{2} + \displaystyle\displaystyle\frac{{n(n + 1)}}{2}} \right] = \displaystyle\frac{{n({n^2} + 1)}}{2}$
So ${S_{50}}$ = $\displaystyle\frac{{50({{50}^2} + 1)}}{2} = 25 \times 2501$ = 62625
5. Find the value of $\displaystyle\frac{2}{3} + \frac{{16}}{{{3^2}}} + \frac{{78}}{{{3^3}}} + \frac{{320}}{{{3^4}}} + \frac{{1210}}{{{3^5}}} + ..........$
Sol:
$\displaystyle\frac{2}{3} + \frac{{16}}{{{3^2}}} + \frac{{78}}{{{3^3}}} + \frac{{320}}{{{3^4}}} + \frac{{1210}}{{{3^5}}} + ..........$ = $1\left( {1 - \displaystyle\frac{1}{3}} \right) + 2\left( {1 - \displaystyle\frac{1}{{{3^2}}}} \right) + 3\left( {1 - \displaystyle\frac{1}{{{3^3}}}} \right) + ..........$
(1 + 2 + 3 +........10) - ($\displaystyle\frac{1}{3} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ......$
[$\displaystyle\frac{1}{3} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ......$ is a Arithmetico geometric progression
${S_\infty }$ = $\displaystyle\frac{a}{{1 - r}} + \frac{{dr}}{{{{\left( {1 - r} \right)}^2}}}$ for AGP ]
Here a = 1/3; d = 1; r = 1/3
$\displaystyle\frac{1}{3} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ......$ = $\displaystyle\frac{{\displaystyle\frac{1}{3}}}{{1 - \displaystyle\frac{1}{3}}} + \displaystyle\frac{{1.\displaystyle\frac{1}{3}}}{{{{\left( {1 - \displaystyle\frac{1}{3}} \right)}^2}}} = \displaystyle\frac{5}{4}$
$ \Rightarrow $ 55 - $\displaystyle\frac{5}{4} = \displaystyle\frac{{215}}{4}$
1. Find the sum to n terms of series 1.2.3 + 2.3.4 + 3.4.5+ ........
a. (n+1)(n+2)(n+3)/3
b. n(n+1)(2n+2)(n+2)/4
c. n(n+1)(n+2)
d. n(n+1)(n+2)(n+3)/4
Sol:
The general term of the above series = n(n+1)(n+2)
Sum = $\sum {n(n + 1)(n + 2) = \sum {{n^3} + 3{n^2} + 2n = \sum {{n^3} + 3\sum {{n^2}} } } } + 2\sum n $
( Recap the formulas for the sum of first n natural numbers, and its squares and its cubes)
$ \Rightarrow \displaystyle\frac{{{n^2}{{(n + 1)}^2}}}{4} + 3.\displaystyle\frac{{n(n + 1)(2n + 1)}}{6} + 2.\displaystyle\frac{{n(n + 1)}}{2}$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{{n^2} + n + 4n + 2 + 4}}{4}} \right]$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{n(n + 1)}}{4} + \frac{{2n + 1}}{2} + 1} \right]$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{{n^2} + 5n + 6}}{4}} \right]$
$ \Rightarrow n(n + 1)\left[ {\displaystyle\frac{{(n + 2)(n + 3)}}{4}} \right]$
So option 4.
Alternate method:
Above mentioned method to be used only when the sum upto a certain terms was asked. Say upto 20 terms. If infinite terms sum was asked, Simply we check the answer options. Take the sum upto 3 terms.
6 + 24 + 60 = 90.
Now substitute n = 3 in the answer options. Only Option 4 satisfies.
2. $\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} + \sqrt {1 + \displaystyle\frac{1}{{{2^2}}} + \displaystyle\frac{1}{{{3^2}}}} + ..........\sqrt {1 + \displaystyle\frac{1}{{{{2007}^2}}} + \displaystyle\frac{1}{{{{2008}^2}}}} $ =
a. 2008 $ - \displaystyle\frac{1}{{2008}}$
b. 2007 $ - \displaystyle\frac{1}{{2007}}$
c. 2007 $ - \displaystyle\frac{1}{{2008}}$
d. 2008 $ - \displaystyle\frac{1}{{2007}}$
Sol:
The clue to solve the problem lies in the options provided. Observe that there are total 2007 terms given. and options are in 2007's and 2008's
Consider only first term:
$\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} $ = $\sqrt {1 + 1 + \displaystyle\frac{1}{4}} = \sqrt {\displaystyle\frac{9}{4}} = \displaystyle\frac{3}{2} = 2 - \displaystyle\frac{1}{2}$
Consider first two terms:
$\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} + \sqrt {1 + \displaystyle\frac{1}{{{2^2}}} +\displaystyle \frac{1}{{{3^2}}}} $
$ \Rightarrow $ $\displaystyle\frac{3}{2} + \sqrt {1 + \frac{1}{4} + \frac{1}{9}} = \frac{3}{2} + \sqrt {\frac{{49}}{{36}}} = \frac{3}{2} + \frac{7}{6} = \frac{8}{3} = 3 - \frac{1}{3}$
Similarly: $\sqrt {1 + \displaystyle\frac{1}{{{1^2}}} + \displaystyle\frac{1}{{{2^2}}}} + \sqrt {1 + \displaystyle\frac{1}{{{2^2}}} + \displaystyle\frac{1}{{{3^2}}}} + ..........\sqrt {1 + \displaystyle\frac{1}{{{{2007}^2}}} + \displaystyle\frac{1}{{{{2008}^2}}}} $ = 2008 - $\displaystyle\frac{1}{{2008}}$
3. S= $\displaystyle\frac{1}{{1 + {1^2}}} + \frac{1}{{2 + {2^2}}} + \frac{1}{{3 + {3^2}}} + ........$ What is the value of S?
(1) 2
(2) 1.5
(3) 1
(4) The sum is not finite.
Sol: Give S= $\displaystyle\frac{1}{{1 + {1^2}}} + \frac{1}{{2 + {2^2}}} + \frac{1}{{3 + {3^2}}} + ........$
$ \Rightarrow \displaystyle\frac{1}{{1(1 + 1)}} + \frac{1}{{2(1 + 2)}} + \frac{1}{{3(1 + 3)}} + ........$
$ \Rightarrow \displaystyle\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ........$
$ \Rightarrow \left( {\displaystyle\frac{1}{1} - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ........$
$ \Rightarrow \displaystyle\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + .......$
4. N, the set of natural numbers, is partitioned into subsets ${S_{_1}}$ = {1}, ${S_{_2}}$ = {2, 3}, ${S_{_3}}$={4, 5, 6}, ${S_{_4}}$ = {7, 8, 9, 10} ...... Find the sum of the elements in the subset ${S_{50}}$
Sol: Observe that the last term of every subset is 1, 1 +2, 1+2+3, 1+2+3+4, ....
So last term is written as $\displaystyle\frac{{n(n + 1)}}{2}$
Now nth subset has n elements in Arithmetic Progression with common difference 1.
To get the first term we have to substract (n-1) from the last term. (take S4, if we substract 3 from 7, we get the first term of that subset)
So first term = $\displaystyle\frac{{n(n + 1)}}{2} - (n - 1) = \displaystyle\frac{{{n^2} + n - 2n + 2}}{2} = \displaystyle\frac{{{n^2} - n + 2}}{2}$
So sum = $\displaystyle\frac{n}{2}\left[ {\displaystyle\frac{{{n^2} - n + 2}}{2} + \displaystyle\displaystyle\frac{{n(n + 1)}}{2}} \right] = \displaystyle\frac{{n({n^2} + 1)}}{2}$
So ${S_{50}}$ = $\displaystyle\frac{{50({{50}^2} + 1)}}{2} = 25 \times 2501$ = 62625
5. Find the value of $\displaystyle\frac{2}{3} + \frac{{16}}{{{3^2}}} + \frac{{78}}{{{3^3}}} + \frac{{320}}{{{3^4}}} + \frac{{1210}}{{{3^5}}} + ..........$
Sol:
$\displaystyle\frac{2}{3} + \frac{{16}}{{{3^2}}} + \frac{{78}}{{{3^3}}} + \frac{{320}}{{{3^4}}} + \frac{{1210}}{{{3^5}}} + ..........$ = $1\left( {1 - \displaystyle\frac{1}{3}} \right) + 2\left( {1 - \displaystyle\frac{1}{{{3^2}}}} \right) + 3\left( {1 - \displaystyle\frac{1}{{{3^3}}}} \right) + ..........$
(1 + 2 + 3 +........10) - ($\displaystyle\frac{1}{3} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ......$
[$\displaystyle\frac{1}{3} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ......$ is a Arithmetico geometric progression
${S_\infty }$ = $\displaystyle\frac{a}{{1 - r}} + \frac{{dr}}{{{{\left( {1 - r} \right)}^2}}}$ for AGP ]
Here a = 1/3; d = 1; r = 1/3
$\displaystyle\frac{1}{3} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ......$ = $\displaystyle\frac{{\displaystyle\frac{1}{3}}}{{1 - \displaystyle\frac{1}{3}}} + \displaystyle\frac{{1.\displaystyle\frac{1}{3}}}{{{{\left( {1 - \displaystyle\frac{1}{3}} \right)}^2}}} = \displaystyle\frac{5}{4}$
$ \Rightarrow $ 55 - $\displaystyle\frac{5}{4} = \displaystyle\frac{{215}}{4}$
