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)r^{3} = (sum of the cubes of the first n natural numbers)
(iv) r^{4} = (n +^{ }1)^{ }(2n^{ }+^{ }1)^{ }(3n²^{ }+^{ }3n^{ }^{ }1)
Method Of Differences :
Let u_{1}, u_{2}, u_{3}...... be a sequence, such that u_{2}  u_{1}, u_{3}  u_{2},............. is either an A.P. or a G.P. then nth term u_{n} of this sequence is obtained as follows
S = u_{1} + u_{2} + u_{3} + ....... + u_{n} ...(i)
S = u_{1} + u_{2} + .......... + u_{n  1} + u_{n} ...(ii)
(i)  (ii) ⇒u_{n} = u_{1} + (u_{2}  u_{1}) + (u_{3}  u_{2}) + ........... + (u_{n}  u_{n  1})
Where the series (u_{2}  u_{1}) + (u_{3}  u_{2}) + ......... + (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. u_{2}u_{1}, u_{3}u_{2} ... u_{n}u_{n  1}
is always either in A.P. or in G.P. in such case let u_{1} = T_{1}, u_{2}  u_{1} = T_{2}, u_{3}  u_{2} = T_{3} .......,u_{n}  u_{n  1} = T_{n}.
So u_{n} = T_{1} + T_{2} + ....... + T_{n} .... (i)
u_{n} = T_{1} + T_{2} + ........ + T_{n 1} + T_{n} .... (ii)
(i)  (ii) ⇒ T_{n} = T_{1} + (T_{2}  T_{1}) + (T_{3}  T_{2}) + ...... + (T_{n} + T_{n  1})
Now, the series (T_{2}  T_{1}) + (T_{3}  T_{2}) + ..... + (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) + ^{ }((10^{2 }^{ }1) + (10^{3} ^{ }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(2^{n}  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)]
= [n^{2} + 3n + 26].
Ex.38 If S_{1}, S_{2}, S_{3}, ... 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}
∴ =
= 2^{2} + 3^{2} + 4^{2} + ....... +(2n  1)^{2} + (2n)^{2} = 1^{2} + 2^{2} + 3^{2} + ..... + (2n)^{2}  1
= sum of the square of the first (2n) natural numbers =1
=
Ex.39 Find the sum to nterms 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) = n^{2} + n + 1
Hence S = S(n^{2} + n + 1) = S n^{2} + S n + S 1 = + n = (n^{2} + 3n + 5)
Ex.40 Find the sum of nterms 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 [3n1] = T_{n}  T_{n  1} = 3n^{2} n
∴ general term of given series is
Hence sum of this series is
S = = = (3n^{2} + 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_{n1}  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(2^{n}  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 a_{1} = 1. Find an equation expressing a_{n} as a polynomial in n. Also find the sum to n terms of the sequence.
Sol. a_{1} = 1 ; a_{2} = a_{1} + 4 = 1 + 4 = 5 ; a_{3} = a_{2} + 6 = 5 + 6 = 11 ; a_{4} = a_{3} + 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) = n^{2} + n  1
sum = = +  =+  n [(n + 1)(2n + 1) + 3(n + 1)  6]
= [2n^{2} + 3n + 1 + 3n + 3  6] = [2n^{2} + 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} = n^{2}  3n + 3
then T_{n} = St_{n} = Sn^{2}  3 S n + 3 S 1 =
= {2n^{2} + 3n + 1  9n  9 + 18} T_{n} = (2n^{2}  6n + 10)= (n^{2}  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 T_{4} = 6a + 6b + 3c + d = 12
a = , b = 1, c = 1, d = 1 Hence T_{n} = (n^{2}  3n + 5)
Ex.45 The squares of the natural numbers are grouped like (1^{2}); (2^{2}, 3^{2}, 4^{2}); (5^{2}, 6^{2}, 7^{2}, 8^{2}, 9^{2}); and so on. Find the sum of the elements in n^{th} group.
Sol. By observations, the last element of the n^{th} group = (n^{2})^{2}
The number of elements in n^{th} group = (2n  1)
The first element of the n^{th} group = (n^{2}  2n + 2)^{2}
Hence sum of the numbers in n^{th} group are
S = (n^{2}  2n + 2)^{2} + (n^{2}  2n + 3)^{2} + ..... + (n^{2})^{2}
⇒ S = [1^{2} + 2^{2} + 3^{2} + ..... (n^{2})^{2}]  [1^{2} + 2^{2} + 3^{2} + ((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} {2^{n} + 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
75 videos238 docs91 tests

1. What is a summation of series? 
2. How do you calculate the sum of an arithmetic series? 
3. What is the formula for finding the sum of a geometric series? 
4. Can the sum of an infinite series be calculated? 
5. Are there any other methods to calculate the sum of a series? 

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