NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.4
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1. Find the sum of the following series upto n terms:
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + .....
Answer
Let Tn denotes the nth term of the given series
Tn = [nth term of 1, 2, 3, ......] × [nth term of 2, 3, 4, .....]
= [1 + (n – 1) 1] [2 + (n – 1)1]
= n (n + 1)
Tn = n 2 + n
2. Find the sum of the following series upto n terms: 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + .......
Answer
Let Tn denote the nth term of the given series.
Then
Tn = [nth term of 1, 2, 3, ......] [n th term of 2, 3, 4, .....] [n th term of 3, 4, 5, .....]
= [1 + (n – 1)1] [2 + (n – 1).1] [3 + (n –1).1]
= n (n + 1) (n + 2)
= n(n 2 + 2n + 3) = n 3 + 3n 2 + 2n
∴ Sn = ∑n3 + 3∑n2 + 2∑n
= (n2(n + 1)2)/4 + (3n(n + 1)(2n +1))/6 + 2.(n(n+1))/2
= (n(n + 1))/4 [n(n + 1) + 2(2n + 1) + 4]
= (n(n + 1))/4 [n2 + n + 4n + 2 + 4]
= (n(n + 1))/4 (n2 + 5n + 6)
= (n(n + 1)(n + 2)(n + 3))/4
3. Find the sum of n terms of the following series.
3 × 12 + 5 × 22 + 7 × 32 + .....
Answer
Let Tn denote the nth term of the given series.
Then,
Tn = [nth term of 3, 5, 7,...] [nth term of 1, 2, 3, ...]2
= {3 + 2 (n – 1)2}{1 + (n – 1).1]2
= (2n + 1) (n)2
= n 2 (2n + 1) = 2n 3 + n 2
Sn = 2∑n3 + ∑n2
= 2. (n2(n + 1)2)/4 + (n(n + 1)(2n + 1))/6
= (n(n + 1))/6 . [3n(n + 1) + (2n + 1)]
= (n(n + 1))/6 (3n2 + 3n + 2n + 1)
= (n(n+1)(3n2 + 5n + 1))/6
4. Find the sum of n terms of the following series. 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + …………
Answer
Tn = 1/([nth term of 1, 2, 3, …..][n th term of 2, 3, 4, ….])
= 1/([1 + (n -1)1][2 + (n – 1)1]) = 1/(n(n + 1))
Let 1/(n(n + 1)) = A/n + B/(n + 1) ……….(1)
=> 1 = A(n + 1) + Bn …………(2)
[on multiplying both sides by n(n + 1)]
To find A: Put n = 0 in (2), we get
1 = A(0 + 1) => A = 1
To find B: put n = -1 in (2), we get
1 = B(-1) => B = -1
Put these values of A and B in (1), the partial fractions are
1/(n(n + 1)) = 1/n – 1/(n + 1)
∴ Tn = 1/n – 1/(n + 1)
Putting n = 1, 2, 3, ………n, we get,
T1 = 1/1 – 1/2
T2 = 1/2 – 1/3
T3 = 1/3 – 1/4
……………………..
Tn = 1/n – 1/(n + 1)
Adding vertically, we get,
Sn = 1 – 1/(n + 1) = (n + 1 – 1)/(n + 1) = n/(n + 1)
5. Find the sum of the following series upto n terms:
5 2+ 62 + 72 +.... 202
Tn = [nth term of 1, 2, 3, ......] × [nth term of 2, 3, 4, .....]
= [1 + (n – 1) 1] [2 + (n – 1)1]
= n (n + 1)
Tn = n 2 + n
= (n(n +1)(2n + 4))/6
= (2n(n + 1)(n + 2))/6 = (n(n + 1)(n + 2))/3
2. Find the sum of the following series upto n terms: 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + .......
Answer
Let Tn denote the nth term of the given series.
Then
Tn = [nth term of 1, 2, 3, ......] [n th term of 2, 3, 4, .....] [n th term of 3, 4, 5, .....]
= [1 + (n – 1)1] [2 + (n – 1).1] [3 + (n –1).1]
= n (n + 1) (n + 2)
= n(n 2 + 2n + 3) = n 3 + 3n 2 + 2n
∴ Sn = ∑n3 + 3∑n2 + 2∑n
= (n2(n + 1)2)/4 + (3n(n + 1)(2n +1))/6 + 2.(n(n+1))/2
= (n(n + 1))/4 [n(n + 1) + 2(2n + 1) + 4]
= (n(n + 1))/4 [n2 + n + 4n + 2 + 4]
= (n(n + 1))/4 (n2 + 5n + 6)
= (n(n + 1)(n + 2)(n + 3))/4
3. Find the sum of n terms of the following series.
3 × 12 + 5 × 22 + 7 × 32 + .....
Answer
Let Tn denote the nth term of the given series.
Then,
Tn = [nth term of 3, 5, 7,...] [nth term of 1, 2, 3, ...]2
= {3 + 2 (n – 1)2}{1 + (n – 1).1]2
= (2n + 1) (n)2
= n 2 (2n + 1) = 2n 3 + n 2
Sn = 2∑n3 + ∑n2
= 2. (n2(n + 1)2)/4 + (n(n + 1)(2n + 1))/6
= (n(n + 1))/6 . [3n(n + 1) + (2n + 1)]
= (n(n + 1))/6 (3n2 + 3n + 2n + 1)
= (n(n+1)(3n2 + 5n + 1))/6
4. Find the sum of n terms of the following series. 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + …………
Answer
Tn = 1/([nth term of 1, 2, 3, …..][n th term of 2, 3, 4, ….])
= 1/([1 + (n -1)1][2 + (n – 1)1]) = 1/(n(n + 1))
Let 1/(n(n + 1)) = A/n + B/(n + 1) ……….(1)
=> 1 = A(n + 1) + Bn …………(2)
[on multiplying both sides by n(n + 1)]
To find A: Put n = 0 in (2), we get
1 = A(0 + 1) => A = 1
To find B: put n = -1 in (2), we get
1 = B(-1) => B = -1
Put these values of A and B in (1), the partial fractions are
1/(n(n + 1)) = 1/n – 1/(n + 1)
∴ Tn = 1/n – 1/(n + 1)
Putting n = 1, 2, 3, ………n, we get,
T1 = 1/1 – 1/2
T2 = 1/2 – 1/3
T3 = 1/3 – 1/4
……………………..
Tn = 1/n – 1/(n + 1)
Adding vertically, we get,
Sn = 1 – 1/(n + 1) = (n + 1 – 1)/(n + 1) = n/(n + 1)
5. Find the sum of the following series upto n terms:
5 2+ 62 + 72 +.... 202
Answer
Tn of given series
= (nth term of 5, 6, 7, ....)2
= [5 + (n + 1) × 1]2
= (n + 4)2 = n 2 + 8n + 16
Sn = ∑n2 + 8∑n + 16 × n
= (n(n + 1)(2n + 1))/6 + 8 × (n(n + 1))/2 + 16n
= n/6[(n + 1)(2n + 1) + 24(n + 1) + 96]
= n/6[2n2 + n + 2n + 1+ 24n + 24 + 96]
= n/6 (2n2 + 27n + 121)
Put n = 16, then
S = 16/6 ( 2 × 162 + 27 × 16 + 121)
8/3 (512 + 432 + 121) = 8/3 × 1065
= 8 × 355 = 2840
Let Tn = 20, a = 5
d = 1
=> 20 = 5 + (n –1).1
or 20 = 5 + n – 1
or 20 – 4 = n
or n = 16
6. Find the sum of the following series upto n terms
3 × 8 + 6 × 11 + 9 × 14 +....
Answer
Here the series is formed by multiplying the corresponding terms of two series both of which are A.P.
viz. 3, 6, 9, ...... and 8, 11, 14, ......
Tn of given series = (nth term of 3, 6, 9,...) × (nth term of 8, 11, 14,...)
= [3 + (n –1)3] [8 + (n – 1)3]
= [3n] [3n + 5] = 9n 2 + 15n
Sn = 9∑n2 + 15 ∑n
= 9 × (n(n + 1)(2n + 1))/6 + (15.n(n + 1))/2
= 3/2 . n(n + 1)[2n + 1 + 5]
= 3/2 . n(n + 1)(2n + 6) = 3n(n + 1)(n + 3)
7. Find the sum to n terms of given series, 1 2 + (12 + 22) + (12 + 22 + 32) + ................
Answer
Let Tn denote the n th term, then
Tn = 12 + 22 + 32 + ..........+ n 2
= ∑n2 = (n(n + 1)(2n + 1))/6
= 1/6 (2n3 +3n2 + n)
Sn = 1/6 [2 ∑n3 + 3 ∑n2 + ∑n]
= 1/6 [ 2. (n2 (n + 1)2)/4 + 3. (n(n +1 )(2n + 1))/6 + (n(n + 1))/2]
= (n(n + 1))/12 [n(n + 1) + (2n + 1) + 1]
= (n(n + 1))/12 [n2 + n + 2n + 1+ 1]
= (n(n + 1)(n2 + 3n + 2))/12 = (n(n + 1)(n + 1)(n + 2))/12
= (n(n + 1)2(n + 2))/12
8. Find the sum to n terms of the following series whose nth term is given by:
n(n + 1) (n + 4)
Answer
Tn = n(n + 1) (n + 2) = n(n 2 + 5n + 4)
= n 3 + 5n 2 + 4n
Sn =
= [(n(n + 1)2)/2] + 5. (n(n + 1)(2n + 1))/6 + 4 . (n(n + 1))/2
= (n2(n + 1)2)/4 + (5n(n + 1)(2n + 1))/6 + 2n(n + 1)
= (n(n + 1))/12 [3n(n + 1) + 10(2n + 1) + 2n]
= (n(n + 1))/12 [3n2 + 3n + 20n + 10 + 2n]
= (n(n + 1))/12 [3n2 + 23n + 3n]
= 1/12 . n(n + 1)[3n2 + 23n + 3n]
9. Find the sum to n terms of the following series whose n th term is given by:
n 2 + 2n
Answer
Tn = n 2 + 2n
Put n = 1, 2, 3, ....., n, we get,
T1 = 12 + 21
T2 = 22 + 22
T3 = 32 + 23
.........................
Tn = n 2 + 2n
Adding vertically, we have,
Sn = (12 + 22 + 32 +... + n 2) + (21 + 22 + 23 +... + 2n)
= ∑n2 + (2(2n – 1))/(2 - 1)
= (n(n + 1)(2n + 1))/6 + 2(2n – 1)
10. Find the sum to n terms of the given series whose n th term is (2n –1)2.
Answer
Tn = (2n – 1)2 = 4n 2 – 4n + 1
Sn =
= (4n(n + 1)(2n + 1))/6 - (4n(n + 1))/2 + n
= n/3 [2(n + 1)(2n + 1) – 6(n + 1) + 3]
= n/3 [2(2n2 + 3n + 1) – 6n – 6 + 3]
Sn = n/3 [4n2 – 1] = (n(2n – 1)(2n + 1))/3