Question (1)
Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ....
Solution
Let us consder 1, 2, 3, 4, .....
It is an A.P. with a = 1, d= 1
i
th term = a+(i-1)d = 1 +(-i1)1 = i
Now 2, 3, 4, 5, ...
It is an A.P. with a = 2, d = 1
i
th term a
i = a +(i-1)d = 2 +(i-1)1 = i+1
∴ i
th term of series
a
i = i(i+1)
= i
2 + i
\[ \text{Now}{S_n} = \sum\limits_{i = 1}^n {{a_i}} \]
\[{S_n} = \sum\limits_{i = 1}^n {\left( {{i^2} + i} \right)} \]
\[{S_n} = \sum\limits_{i = 1}^n {{i^2}} + \sum\limits_{i = 1}^n i \]
\[{S_n} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2}\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{6}\left[ {2n + 1 + 3} \right]\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{6}\left[ {2n + 4} \right]\]
\[{S_n} = \frac{{n\left( {n + 1} \right)2\left( {n + 2} \right)}}{6} = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}\]
Question (2)
Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Solution
Let us consider 1
st nos = 1, 2, 3, ...
It is A.P. whith a = 1, d = 1, i
th a
i = a + (i-1) d
a
i = 1 + (i-1)(1) = i
Now 2
nd nos = 2, 3, 4 ...
It is A.P. with a = 2, d = 1 so a
i = a + (i-1)d
a
i = 2 + (i-1)1 = i+1
Now 3
rd nos = 3, 4, 5, ...
It is an A.P. with a = 3, d = 1
So a
i = a + (i-1)d = 3 +(i-1)1 = i+2
So i
th term of series
a
i = i(i+1)(i+2)
a
i = i(i
2 +3i +2)
a
i = i
3 +3i
2 +2i
\[{S_n} = \sum\limits_{i = 1}^n {{a_i}} \]
\[{S_n} = \sum\limits_{i = 1}^n {\left( {{i^3} + 3{i^2} + 2i} \right)} \]
\[{S_n} = \sum\limits_{i = 1}^n {{i^3} + 3\sum\limits_{i = 1}^n {{i^2} + 2\sum\limits_{i = 1}^n i } } \]
\[ = \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + 3\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 2\frac{{\left( {n + 1} \right)}}{2}\]
\[ = \frac{{n\left( {n + 1} \right)}}{4}\left[ {n\left( {n + 1} \right) + 2\left( {2n + 1} \right) + 4} \right]\]
\[ = \frac{{n\left( {n + 1} \right)}}{4}\left[ {{n^2} + n + 4{n} + 2 + 4} \right]\]
\[ = \frac{{n\left( {n + 1} \right)}}{4}\left[ {{n^2} + 5n + 6} \right]\]
\[ = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{4}\]
Question (3)
Find the sum to n terms of the series 3 × 1
2 + 5 × 2
2 + 7 × 3
2 + ...
Solution
The given series is 3 × 1
2 + 5 × 2
2 + 7 × 3
2 + ...
It is an A.P. with a=3, d = 2
It is an A.P. with a = 3, d = 2
a
i = a+(i-1)d = 3+(i-1)2 =2i+1
Now 2
nd nos = 1
2, 2
2, 3
2, ....
a
i = i
2
So i
th term of series
a
i = (2i+1)i
2 = 2i
3 + i
2
\[ \text{Now} \;{S_n} = \sum {{a_i}} \]
\[{S_n} = \sum {\left( {2{i^3} + {i^2}} \right)} \]
\[{S_n} = 2\sum {{i^3}} + \sum {{i^2}} \]
\[{S_n} = 2\frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{6}\left[ {3n\left( {n + 1} \right) + 2n + 1} \right]\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{6}\left( {3{n^2} + 3n + 2n + 1} \right)\]
\[{S_n} = \frac{{n\left( {n + 1} \right)\left( {3{n^2} + 5n + 1} \right)}}{6}\]
Question (4)
Find the sum to n terms of the series $\frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}}$
Solution
Now 1
st we take 1
st number of denominater
=1 ,2 ,3, ... . it is an A.P. with a = 1, d = 1
a
i = a + (i-1)d = 1 +(i-1)1 = i
Now 2
nd numbers = 2, 3, 4, ...
It is an A.P. i =2, d = 1
a
i = a +(i -1 ) 1 = 2 +(i-1) = i + 1
Now i
th term of series
${a_i} = \frac{1}{{i\left( {i + 1} \right)}}$
${S_n} = \sum\limits_{i = 1}^n {{a_i}} $
\[{S_n} = \sum\limits_{i = 1}^n {\frac{1}{{i\left( {i + 1} \right)}}} \]
\[{S_n} = \sum\limits_{i = 1}^n {\frac{{\left( {i + 1} \right) - \left( i \right)}}{{i\left( {i + 1} \right)}}} \]
\[{S_n} = \sum\limits_{i = 1}^n {\frac{{\left( {i + 1} \right)}}{{i\left( {i + 1} \right)}} - } \sum\limits_{i = 1}^n {\frac{i}{{i\left( {i + 1} \right)}}} \]
\[{S_n} = \sum\limits_{i = 1}^n {\frac{1}{i} - } \sum\limits_{i = 1}^n {\frac{1}{{\left( {i - 1} \right)}}} \]
\[{S_n} = \left[ {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}} \right] - \left[ {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{n + 1}}} \right]\]
\[{S_n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} - \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{n + 1}}\]
\[{S_n} = 1 - \frac{1}{{n + 1}}\]
\[{S_n} = \frac{n}{{n + 1}}\]
Question (5)
Find the sum to n terms of the series 5
2 + 6
2 + 7
2 ....20
2
Solution
series 5
2 + 6
2 + 7
2 ....20
2
=(1
2 + 2
2 + ...+20
2) - (1
2 + 2
2 + 3
2 + 4
2)
\[{S_n} = \sum\limits_{i = 1}^{20} {{i^2} - \sum\limits_{i = 1}^4 {{i^2}} } \]
\[{S_n} = \frac{{20\left( {20 + 1} \right)\left( {40 + 1} \right)}}{6} - \frac{{4\left( {4 + 1} \right)\left( {8 + 1} \right)}}{6}\]
\[{S_n} = \frac{{20\left( {21} \right)\left( {41} \right)}}{6} - \frac{{4\left( 5 \right)\left( 9 \right)}}{6}\]
=2870 - 30 = 2840
Question (6)
Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +...
Solution
3 × 8 + 6 × 11 + 9 × 14 +...
1
st number 3, 6, 9, ...
It is an A.P. with a = 3, d = 3
a
i = a + (i-1)d = 3 + (i-1)3 = 3i
2
nd numbers are 8, 11, 14, ....
It is an A.P. with a = 8 and d = 3
a
i = a+(i-1)d
a
i = 8 + (i - 1 )3
a
i = 3i + 5
so i
th term
a
i = 3i (3i+5) = 9i
2 + 15i
$ \text{Now}\;{S_n} = \sum\limits_{i = 1}^n {{a_i}} $
${S_n} = \sum\limits_{i = 1}^n {9{i^2} + 15i} $
\[{S_n} = \sum\limits_{i = 1}^n {9{i^2} + 15\sum\limits_{i = 1}^n i } \]
\[{S_n} = \frac{{9n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{15n\left( {n + 1} \right)}}{2}\]
\[{S_n} = \frac{{3n\left( {n + 1} \right)}}{2}\left[ {2n + 1 + 5} \right]\]
\[{S_n} = \frac{{3n\left( {n + 1} \right)}}{2}\left( {2n + 6} \right)\]
\[{S_n} = \frac{{3n\left( {n + 1} \right)}}{2} \cdot 2\left( {n + 3} \right)\]
\[{S_n} = 3n\left( {n + 1} \right)\left( {n + 3} \right)\]
Question (7)
Find the sum to n terms of the series 1
2 + (1
2 + 2
2) + (1
2 + 2
2 + 3
2) + …
Solution
The given series is
1
2 + (1
2 + 2
2) + (1
2 + 2
2 + 3
2) + …
i
th term = 1
2 + 2
2 + 3
2 + .....i
2
\[{i^{th} \; \text{term}} = {1^2} + {2^2} + {3^2} + ....{i^2}\]
\[{i^{th}\; \text{term}} = \sum\limits_{i = 1}^r {{i^2}} \]
\[{i^{st}\; \text{term}} = \frac{{r\left( {r + 1} \right)\left( {2r + 1} \right)}}{6}\]
\[ = \frac{{r\left( {2{r^2} + 3r + 1} \right)}}{6}\]
\[ = \frac{{2{r^3} + 3{r^2} + r}}{6}\]
\[{S_n} = \sum\limits_{i = 1}^n {{a_i}} \]
\[{S_n} = \sum\limits_{i = 1}^n {\frac{{2{r^3} + 3{r^2} + r}}{6}} \]
\[{S_n} = \frac{2}{6}\sum\limits_{i = 1}^n {{r^3} + \frac{3}{6}\sum\limits_{i = 1}^n {{r^2} + \frac{1}{6}\sum\limits_{i = 1}^n r } } \]
\[{S_n} = \frac{2}{6}\frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{1}{2}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{1}{6}\frac{{n\left( {n + 1} \right)}}{2}\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {n\left( {n + 1} \right) + 2n + 1 + 1} \right]\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {{n^2} + n + 2n + 2} \right]\]
\[{S_n} = \frac{{n\left( {n + 1} \right)\left( {{n^2} + 3n + 2} \right)}}{{12}}\]
\[{S_n} = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 1} \right)}}{{12}} = \frac{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}{{12}}\]
Question (8)
Find the sum to n terms of the series whose n
th term is given by n (n + 1) (n + 4).
Solution
n(n+1)(n+4) = n(n
2 + 5n +4) = n
3 +5n
2 +4n
\[{S_n} = \sum {\left( {{n^3} + 5{n^2} + 4n} \right)} \]
\[{S_n} = \sum {{n^3} + 5\sum {{n^2} + 4\sum n } } \]
\[{S_n} = \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{{5n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{4n\left( {n + 1} \right)}}{2}\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {3n\left( {n + 1} \right) + 10\left( {2n + 1} \right) + 24} \right]\]
\[{S_n} = \frac{{n\left( {n + 1} \right)}}{{12}}\left( {3{n^2} + 3n + 20n + 10 + 24} \right)\]
\[{S_n} = \frac{{n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)}}{{12}}\]
Question (9)
Find the sum to n terms of the series whose n
th terms is given by n
2 + 2
n
Solution
n
2 + 2
n
\[{S_n} = \sum {\left( {{n^2} + {2^n}} \right)} \]
\[{S_n} = \sum {{n^2} + \sum {{2^n}} } \]
\[{S_n} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \sum {{2^n}} \]
\[\sum {{2^n}} = 2 + {2^2} + {2^3} + ...\]
It is G.P. with a = 2, r = 2 > 1
\[{S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} = \frac{{2\left( {{2^n} - 1} \right)}}{1} = 2\left( {{2^n} - 1} \right)\]
\[{S_n} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 2\left( {{2^n} - 1} \right)\]
Question (10)
Find the sum to n terms of the series whose n
th terms is given by (2n – 1)
2
Solution
\[{S_n} = \sum {{{\left( {2n - 1} \right)}^2}} \]
\[{S_n} = \sum {\left( {4{n^2} - 4n + 1} \right)} \]
\[{S_n} = 4\sum {{n^2} - 4\sum n } + \sum 1 \]
\[{S_n} = \frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{4n\left( {n + 1} \right)}}{2} + n\]
\[{S_n} = \frac{n}{3}\left[ {2\left( {n + 1} \right)\left( {2n + 1} \right) - 6\left( {n + 1} \right) + 3} \right]\]
\[{S_n} = \frac{n}{3}\left[ {4{n^2} + 6n + 2 - 6n - 6 + 3} \right]\]
\[{S_n} = \frac{n}{3}\left( {4{n^2} - 1} \right)\]
\[{S_n} = \frac{n}{3}\left( {2n - 1} \right)\left( {2n + 1} \right)\]