Today our topic of discussion is Determination of the sum of first n terms of natural numbers.
Determination of the sum of first n terms of natural numbers
Determination of the sum of first n terms of natural numbers
Let Sn be the sum of n-numbers of natural numbers i.e.
Sn =1+2+3+***+ (n – 1) + n
Writing from the first term and conversely from the last term of the series we get,
Sn =1+2+3+***+ (n – 2) + (n – 1) +n…(1)
and Sn+ (n-1) + (n-2)++3+2+1…(2)
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Adding, 2Sn= (n + 1) + (n + 1) + (n + 1) + ··· +(n+1) [n-number of terms]
or, 2Sn = n(n + 1)
therefore Sn = n(n + 1))/2 ..(3)
Example 3. Find the sum total of first 50 natural numbers.
Solution: Using formula (3) we get,
S_{50} = (50(50 + 1))/2 = 25 * 51 = 1275
.. The sum total of first 50 natural numbers is 1275.
Example 4. 1+2+3+4+. +99= what?
Solution:
The first term of the series a = 1 common difference d = 2 – 1 = 1 and the last term p = 99
∴ It is an arithmetic series.
Let the nth term of the series = 9
We know, nth term of an arithmetic progression = a + (n – 1) * d
a + (n – 1) * d = 99
or, 1 + (n – 1) * 1 = 99
or, 1 + n – 1 = 99
∴ n = 99
From (4) formula, the sum of first n-terms of an arithmetic series,
S{n} = n/2 * {2a + (n – 1) * d}
Hence, the sum of first 99 terms of the series S{99} = 99/2 * {2 * 1 + (99 – 1) * 1}
=99/2 * (2 + 98)
= (99 * 100)/2
= 99 * 50
= 4950
Alternative method: From formula (3), S{n} = n/2 * (a + p)
S{99} = 99/2 * (1 + 99) = (99 * 100)/2 = 4950
Example 5. What is the sum of first 30 terms of the series 7 + 12 + 17 +****
Solution: First term of the series a = 7 , common difference d = 12 – 7 = 5
∴It is an arithmetic series. Here, number of terms n = 30 We know that the sum of first n-terms of an arithmetic series,
Sn = n/2 * {2a + (n – 1) * d}
So the sum of 30 terms S{30} = 30/2 * {2 * 7 + (30 – 1) * 5} = 15(14 + 29 * 5)
= 15(14 + 145) = 15 * 159 = 2385
Example 6. Rashid deposits Tk. 1200 from his salary in the first month and every subsequent month, he deposits Tk. 100 more than the previous month.
1) Express the aforesaid problem as a series upto n terms.
2) How much does he deposit in the 18 th month and how much does he deposit in first 18 months?
3) In how many years does he deposit a total of Tk. 106200?
Solution:
1) As per the question, first term of the arithmetic series, a = 1200 common
difference d = 100
second term = 1200+100 = 1300
Third term = 1300 + 100 = 1400
n-th term = a + (n – 1) * (d = 1200 + (n – 1) * 100) = 1100 + 100 ,
∴The series is 1200+1300+1400+***+ (1100 + 100n)
2) We know, n-th term = a + (n – 1) * d
∴ deposit in the 18 th month = a+ (18-1)d=1200+ 17 x 100 = 2900 Tk.
Again, summation of first terms = n/2 * {2a + (n – 1) * d} 18
∴deposit in first 18 months(2 x 1200+ (18-1) x 100) Tk. = 9(2400 + 1700) = 36900Tk .
3) Let, he deposits 106200 Tk. in n months.
According to the question, n/2 * {2a + (n – 1) * d} = 106200
or, n/2 * {2 * 1200 + (n – 1) * 100} = 106200
or, n(2400 + 100n – 100) = 21240C
or, 100n² + 2300n – 212400 = 0
or, n² + 23n – 2124 = 0
or, n² + 59n – 36n – 2124 = 0
or, (n + 59)(n – 36) = 0
So, n = – 59 or, n = 36
Number of months cannot be negative.
∴ The required time: 36 months or 3 years.
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