11th NCERT/CBSE Sequences and series Exercise 9.4 Questions 10
Hi

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
ith term = a+(i-1)d = 1 +(-i1)1 = i
Now 2, 3, 4, 5, ...
It is an A.P. with a = 2, d = 1
ith term ai = a +(i-1)d = 2 +(i-1)1 = i+1
∴ ith term of series
ai = i(i+1)
= i2 + 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 1st nos = 1, 2, 3, ...
It is A.P. whith a = 1, d = 1, ith ai = a + (i-1) d
ai = 1 + (i-1)(1) = i
Now 2nd nos = 2, 3, 4 ...
It is A.P. with a = 2, d = 1 so ai = a + (i-1)d
ai = 2 + (i-1)1 = i+1
Now 3rd nos = 3, 4, 5, ...
It is an A.P. with a = 3, d = 1
So ai = a + (i-1)d = 3 +(i-1)1 = i+2
So ith term of series
ai = i(i+1)(i+2)
ai = i(i2 +3i +2)
ai = i3 +3i2 +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 × 12 + 5 × 22 + 7 × 32 + ...

Solution

The given series is 3 × 12 + 5 × 22 + 7 × 32 + ...
It is an A.P. with a=3, d = 2
It is an A.P. with a = 3, d = 2
ai = a+(i-1)d = 3+(i-1)2 =2i+1
Now 2nd nos = 12, 22, 32, ....
ai = i2
So ith term of series
ai = (2i+1)i2 = 2i3 + i2
$\text{Now} \;{S_n} = \sum {{a_i}}$ ${S_n} = \sum {\left( {2{i^3} + {i^2}} \right)}$ ${S_n} = 2\sum {{i^3}} + \sum {{i^2}}$ ${S_n} = 2\frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ ${S_n} = \frac{{n\left( {n + 1} \right)}}{6}\left[ {3n\left( {n + 1} \right) + 2n + 1} \right]$ ${S_n} = \frac{{n\left( {n + 1} \right)}}{6}\left( {3{n^2} + 3n + 2n + 1} \right)$ ${S_n} = \frac{{n\left( {n + 1} \right)\left( {3{n^2} + 5n + 1} \right)}}{6}$

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 1st we take 1st number of denominater =1 ,2 ,3, ... . it is an A.P. with a = 1, d = 1
ai = a + (i-1)d = 1 +(i-1)1 = i
Now 2nd numbers = 2, 3, 4, ...
It is an A.P. i =2, d = 1
ai = a +(i -1 ) 1 = 2 +(i-1) = i + 1
Now ith 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 52 + 62 + 72 ....202

Solution

series 52 + 62 + 72 ....202
=(12 + 22 + ...+202) - (12 + 22 + 32 + 42) ${S_n} = \sum\limits_{i = 1}^{20} {{i^2} - \sum\limits_{i = 1}^4 {{i^2}} }$ ${S_n} = \frac{{20\left( {20 + 1} \right)\left( {40 + 1} \right)}}{6} - \frac{{4\left( {4 + 1} \right)\left( {8 + 1} \right)}}{6}$ ${S_n} = \frac{{20\left( {21} \right)\left( {41} \right)}}{6} - \frac{{4\left( 5 \right)\left( 9 \right)}}{6}$ =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 +...
1st number 3, 6, 9, ...
It is an A.P. with a = 3, d = 3
ai = a + (i-1)d = 3 + (i-1)3 = 3i
2nd numbers are 8, 11, 14, ....
It is an A.P. with a = 8 and d = 3
ai = a+(i-1)d
ai = 8 + (i - 1 )3
ai = 3i + 5
so ith term
ai = 3i (3i+5) = 9i2 + 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 12 + (12 + 22) + (12 + 22 + 32) + …

Solution

The given series is
12 + (12 + 22) + (12 + 22 + 32) + … ith term = 12 + 22 + 32 + .....i2
${i^{th} \; \text{term}} = {1^2} + {2^2} + {3^2} + ....{i^2}$ ${i^{th}\; \text{term}} = \sum\limits_{i = 1}^r {{i^2}}$ ${i^{st}\; \text{term}} = \frac{{r\left( {r + 1} \right)\left( {2r + 1} \right)}}{6}$ $= \frac{{r\left( {2{r^2} + 3r + 1} \right)}}{6}$ $= \frac{{2{r^3} + 3{r^2} + r}}{6}$ ${S_n} = \sum\limits_{i = 1}^n {{a_i}}$ ${S_n} = \sum\limits_{i = 1}^n {\frac{{2{r^3} + 3{r^2} + r}}{6}}$ ${S_n} = \frac{2}{6}\sum\limits_{i = 1}^n {{r^3} + \frac{3}{6}\sum\limits_{i = 1}^n {{r^2} + \frac{1}{6}\sum\limits_{i = 1}^n r } }$ ${S_n} = \frac{2}{6}\frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{1}{2}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{1}{6}\frac{{n\left( {n + 1} \right)}}{2}$ ${S_n} = \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {n\left( {n + 1} \right) + 2n + 1 + 1} \right]$ ${S_n} = \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {{n^2} + n + 2n + 2} \right]$ ${S_n} = \frac{{n\left( {n + 1} \right)\left( {{n^2} + 3n + 2} \right)}}{{12}}$ ${S_n} = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 1} \right)}}{{12}} = \frac{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}{{12}}$

Question (8)

Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).

Solution

n(n+1)(n+4) = n(n2 + 5n +4) = n3 +5n2 +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 nth terms is given by n2 + 2n

Solution

n2 + 2n
${S_n} = \sum {\left( {{n^2} + {2^n}} \right)}$ ${S_n} = \sum {{n^2} + \sum {{2^n}} }$ ${S_n} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \sum {{2^n}}$ $\sum {{2^n}} = 2 + {2^2} + {2^3} + ...$ 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 nth terms is given by (2n – 1)2

Solution

${S_n} = \sum {{{\left( {2n - 1} \right)}^2}}$ ${S_n} = \sum {\left( {4{n^2} - 4n + 1} \right)}$ ${S_n} = 4\sum {{n^2} - 4\sum n } + \sum 1$ ${S_n} = \frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{4n\left( {n + 1} \right)}}{2} + n$ ${S_n} = \frac{n}{3}\left[ {2\left( {n + 1} \right)\left( {2n + 1} \right) - 6\left( {n + 1} \right) + 3} \right]$ ${S_n} = \frac{n}{3}\left[ {4{n^2} + 6n + 2 - 6n - 6 + 3} \right]$ ${S_n} = \frac{n}{3}\left( {4{n^2} - 1} \right)$ ${S_n} = \frac{n}{3}\left( {2n - 1} \right)\left( {2n + 1} \right)$