Summation of Series | Mathematics (Maths) Class 11 - Commerce PDF Download

G. Summation of series

Remarks :

(i)r =  (sum of the first n natural nos.)

(ii)r² =  (sum of the squares of the first n natural numbers)

(iii)r³ =  (sum of the cubes of the first n natural numbers)

(iv) r⁴ = (n + 1) (2n + 1) (3n² + 3n - 1)

Method Of Differences : 

Let u₁, u₂, u₃...... be a sequence, such that u₂ - u₁, u₃ - u₂,............. is either an A.P. or a G.P. then nth term u_n of this sequence is obtained as follows

S = u₁ + u₂ + u₃ + ....... + u_n ...(i)

S = u₁ + u₂ + .......... + u_{n - 1} + u_n ...(ii)

(i) - (ii) ⇒u_n = u₁ + (u₂ - u₁) + (u₃ - u₂) + ........... + (u_n - u_{n - 1})

Where the series (u₂ - u₁) + (u₃ - u₂) + ......... + (u_n - u_{n - 1}) is

either in A.P. or in G.P. then we can find u_n and hence sum of this series as S =

Remark : It is not always necessary that the series of first order of differences i.e. u₂-u₁, u₃-u₂ ... u_n-u_{n - 1}

is always either in A.P. or in G.P. in such case let u₁ = T₁, u₂ - u₁ = T₂, u₃ - u₂ = T₃ .......,u_n - u_{n - 1} = T_n.
 So u_n = T₁ + T₂ + ....... + T_n .... (i)

u_n = T₁ + T₂ + ........ + T_{n -1} + T_n .... (ii)

(i) - (ii) ⇒ T_n = T₁ + (T₂ - T₁) + (T₃ - T₂) + ...... + (T_n + T_{n - 1})

Now, the series (T₂ - T₁) + (T₃ - T₂) + ..... + (T_n - T_{n - 1}) is series of second order of differences and when it is either in A.P. or in G.P., then u_n =

Otherwise in the similar way we find series of higher order of differences and the n^th term of the series.

If possible express r^th term as difference of two terms as t_r = f(r) - f(r ± 1) . This can be explained with help of examples given below.

Ex.35 Find the sum to  n  terms of the series,

0.7 + 7.7 + 0.77 + 77.7 + 0.777 + 777.7 + 0.7777 + ...... where n is even .

Sol. n = 2m

s = (0.7 + 0.77 + 0.777 + ...... m term) + (7.7 + 77.7 + 777.7 + ...... m terms)

=(0.9 + 0.99 + 0.999 + ..... m terms) +  ((10² - 1) + (10³ - 1) + ..... + (10^{m + 1} - 1))

=  +

Ex.36 Determine the sums of the following series

1.  1 + ; 

2.   1 - .

Sol. To determine the required sums first compute the following sum

For computing the first of the sums put in the deduced formula x = . then we have

1 +  +  +  + .... +  {3(2ⁿ - 1) - 2n}

And putting x = –1/2, we find  

 Ex.37 There are n necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general the ith necklace contains i beads more than the number of beads in (i- 1)th necklace. Find the total number of beads is all the n necklaces.

Sol. Let us write the sequence of the number of beads in the 1st, 2nd, 3rd, ...., nth necklaces.

= 5, 7, 10, 14, 19, ........       = (4 + 1), (4 + 3), (4 + 6) (4 + 10), (4 + 15),. ......,

S_n = Total number of beads in the n necklaces   

S_n =  + 1 + 3 + 6 + .... +

= 4n + Sum of the first n triangular numbers  = 

= 4n +

 = 4n +

=  [48n + 2n(n + 1) (n + 2)]

= [n² + 3n + 26].

Ex.38 If S₁, S₂, S₃, ... S_n, .... are the sums of infinite geometric series whose first terms are 1, 2, 3, ... n, ... and whose common ratios are , , , ....,, ... respectively, then find the value of .

Sol. r^th series will have a = r and common ratio is

∴ S_r =  =  = r + 1

∴  = (r + 1)²

∴  =

= 2² + 3² + 4² + ....... +(2n - 1)² + (2n)² = 1² + 2² + 3² + ..... + (2n)² - 1

= sum of the square of the first (2n) natural numbers =-1 

=

Ex.39 Find the sum to n-terms 3 + 7 + 13 + 21 + ......

Sol. Let S = 3 + 7 + 13 + 21 + ............. + T_n ................(i)

S = 3 + 7 + 13 + ................+ T_{n - 1} + T_n ...............(ii)

(i) - (ii)       ⇒         T_n = 3 + 4 + 6 + 8 + ..................... + (T_n - T_{n - 1})

= 3 + [8 + (n - 2)2] = 3 + (n - 1) (n + 2) = n² + n + 1

Hence S = S(n² + n + 1) = S n² + S n + S 1 =  + n = (n² + 3n + 5)

Ex.40 Find the sum of n-terms 1 + 4 + 10 + 22 + .......

Sol. Let S = 1 + 4 + 10 + 22 + ...... + T_n .... (i)

 S = 1 + 4 + 10 + ........+T_{n - 1} + T_n .... (ii)

(i) - (ii) 

⇒ T_n = 1 + (3 + 6 + 12 + ...... + T_n - T_{n - 1}) 

=

So S =

Ex.41 Find the nth term and the sum of n term of the series 2, 12, 36, 80, 150, 252

Sol. Let S = 2 + 12 + 36 + 80 + 150 + 252 + ........ + T_n    .... (i)

S = 2 + 12 + 36 + 80 + 150 + 252 + ....... + T_{n - 1} + T_n      .... (ii)

(i) = (ii) ⇒ T_n = 2 + 10 + 24 + 44 + 70 + 102 + ...... + (T_n - T_{n - 1})                     .... (iii)

T_n = 2 + 10 + 24 + 44 + 70 + 102 + ......... + (T_{n - 1} - T_{n - 2}) + (T_n - T_{n - 1})          ..... (iv)

(iii) - (iv) ⇒ T_n  - T_{n - 1} = 2 + 8 + 14 + 20 + 26 + .... 

= [4+(n - 1) 6]=n [3n-1] = T_n - T_{n - 1} = 3n² -n

∴ general term of given series is

Hence sum of this series is

S =  =  =  (3n² + 7n + 2)  n (n + 1) (n + 2) (3n + 1)

Ex.42 Find the general term and sum of n terms of the series 9, 16, 29, 54, 103

Sol. Let S = 9 + 16 + 29 + 54 + 103 + .......... + T_n   .... (i)

S = 9 + 16 + 29 + 54 + 103 + ............+ T_{n - 1} + T_n   .... (ii)

(i) - (ii)⇒ T_n = 9 + 7 + 13 + 25 + 49 + .......... + (T_n - T_{n - 1})             .... (iii)

T_n = 9 + 7 + 13 + 25 + 49 + ........ + (T_n-1 - T_{n - 2}) + (T_n - T_{n - 1})            .... (iv)

(iii) - (iv)    ⇒ T_n - T_{n - 1} = 9 + (-2) +  = 7 + 6 [2^{n - 2} - 1] = 6(2)^{n - 2} +1.

∴ General term is T_n = 6(2)^{n - 1} + n + 2

Also sum S = ST_n = 6S2^{n - 1} + Sn + S2 = 6.  + 2n = 6(2ⁿ - 1) +

Ex.43 The n^th term, a_n of a sequence of numbers is given by the formula a_n = a_{n - 1} + 2n for  and a₁ = 1. Find an equation expressing a_n as a polynomial in n. Also find the sum to n terms of the sequence.

Sol. a₁ = 1 ; a₂ = a₁ + 4 = 1 + 4 = 5 ; a₃ = a₂ + 6 = 5 + 6 = 11 ; a₄ = a₃ + 8 = 11 + 8 = 19 and so on

hence S = 1 + 5 + 11 + 19 + ........ + a_n
 S = + 1 + 5 + 11 + ........ + a_{n - 1} + a_n

( - ) ————————————————

0 = 1 + 4 + 6 + 8 + ........ + (a_n - a_{n - 1}) - a_n

a_n = 1 + 2

a_n = 1 + [4 + (n - 2)] = 1 + (n - 1) (n + 2) = n² + n - 1

sum =  =  +  -  =+ - n [(n + 1)(2n + 1) + 3(n + 1) - 6]

= [2n² + 3n + 1 + 3n + 3 - 6] = [2n² + 6n - 2] =

Ex.44 Find the n^th term of the series 1 + 2 + 5 + 12 + 25 + 46 + .....

Sol. Let the sum of the series by S_n and nth term of the series be T_n

Then S_n = 1 + 2 + 5 + 12 + 46 + ...+T_{n -1} + T_n                ...(1)

∴ S_n = 1 + 2 + 5 + 12 + 25 +.... +T_{n - 1} + T_n                 ...(2)

Subtracting (2) from (1), we get 0 = 1 + 1 + 3 + 7 + 13 + 21 + .... + (T_n - T_{n - 1}) - T_n

∴ T_n = 1 + 3 + 7 + 13 + 21 + ....+t_{n - 1} + t_n  ...(3)

(Here nth term of T_n is t_n)

∴ T_n = 1 + 1+ 3 + 7 + 13 + .... t_{n - 1} + t_n ...(4)

Subtracting (4) from (3), we get 0 = 1 + 0 + 2 + 4 + 6 + 8 + .... +(t_n - t_{n - 1}) - t_n

∴ t_n = 1 + 2 + 4 + 6 + 8 +....(n - 1) terms = 1 + (2 + 4 + 6 + 8 (n - 2) terms)

= 1 +  {2.2 + (n - 2 - 1) 2} = 1 + (n - 2) (2 + n - 3)        ∴          t_n = n² - 3n + 3

then T_n = St_n = Sn² - 3 S n + 3 S 1 =

=  {2n² + 3n + 1 - 9n - 9 + 18}  T_n = (2n² - 6n + 10)=  (n² - 3n + 5)

Alternative Method : The nth terms of the series can be written directly from the following procedure

1, 2, 5, 12, 25, 46,......    (given series)

1, 3, 7, 13, 21,.....   (first consecutive differences)

2, 4, 6, 8,....  (second consecutive differences)

2, 2, 2,....    (constant terms)

then T_n = a (n - 1) (n - 2) (n - 3) + b (n - 1) (n - 2) + c(n - 1) + d

Putting n = 1, 2, 3, 4 then we get T₄ = 6a + 6b + 3c + d = 12

 a = , b = 1, c = 1, d = 1 Hence T_n =  (n² - 3n + 5)

Ex.45 The squares of the natural numbers are grouped like (1²); (2², 3², 4²); (5², 6², 7², 8², 9²); and so on. Find the sum of the elements in n^th group.

Sol. By observations, the last element of the n^th group = (n²)²

The number of elements in n^th group = (2n - 1)

The first element of the n^th group = (n² - 2n + 2)²

Hence sum of the numbers in n^th group are

S = (n² - 2n + 2)² + (n² - 2n + 3)² + ..... + (n²)²

⇒ S = [1² + 2² + 3² + ..... (n²)²] - [1² + 2² + 3² + ((n - 1)²)²]

⇒ S =

Ex.46 The natural numbers are arranged in groups as given below ;

 Prove that the sum of the numbers in the nth group is  2^{n - 2} {2ⁿ + 2^{n - 1} - 1} .

Sol . Note that  n^th group has  2^{n - 1} terms .1st term in the nth group is 2^{n - 1} and the last term in the nth group is 2n - 1 .

∴ sum of the terms in the nth group