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Sum of n Terms of Special Series

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Sum of n Terms of Special Series - Lesson Summary

For an AP: a, a + d, a + 2d, a + 3d, a + 4d, ...., a + (n-1)d

Sn = n/2 x [2a + (n - 1)d]

or Sn = n/2 x [a + l], where l = a + (n-1)d

For a G.P: a,ar,ar2,ar3,....,arn-1 …(i)

Sn  =  { na; if r=1 a(1-r ⁿ ) 1-r ; if r<1 a(r ⁿ -1) r-1 ; if r>1 }

1 + 2 + 3 + .... + n, i.e. the sum of first n natural numbers

12 + 22 + 32 + .... + n2, i.e. the sum of the squares of the first n natural numbers

13 + 23 + 33 + .... + n3 , i.e. the sum of the cubes of the first n natural numbers

Sum of the first n natural numbers:

Denote the sum of the first n natural numbers by Sn. Let this be equation 1.

Sn = 1 + 2 + 3 + .... + (n - 1) + n ... (i)

Again, Sn = n + (n - 1) + (n - 2) +  .... + 2 + 1 ... (ii)

Adding both sides of (i) and (ii), we get

2Sn = (1 + n) + (2 + n - 1) + (3 + n - 2) + .... + (n - 1 + 2) + (n + 1)

⇒ 2Sn = (n + 1) + (n + 1) + (n + 1) + --- (n + 1) + (n + 1)  [n terms]

⇒ 2Sn = n(n + 1)

⇒ Sn = n(n + 1) / 2

Alternate method:

Sn = 1 + 2 + 3 + .... + (n - 1) + n ... (i)

Common difference (d) = 1

First term (a) = 1

Last term (l) = n

Sn = n(a + l) / 2

     = n(n + 1) / 2

Hence, the sum of the first n natural numbers is n(n + 1) / 2.

1 + 2 + 3 + ....+ n = ∑ k=1 n k  

Example: Find the sum of the first five natural numbers.

Let S5 = 1 + 2 + 3 + 4 + 5

Then, by the formula, we have S5 = 5(5 + 1) / 2 = 15.

Sum of the squares of the first n natural numbers:

Sn = 12 + 22 + 32 + ... + n2

Now, we use the identity,

k3 - (k - 1)3 = 3k2 - 3k + 1 ... (i)

Substituting k = 1, 2, 3, .... , n successively in (i), we get

13 - (0)3 = 3(1)2 - 3(1) + 1

23 - (1)3 = 3(2)2 - 3(2) + 1

33 - (2)3 = 3(3)2 - 3(3) + 1

…………………………….

…………………………….

n3 - (n - 1)3 = 3(n)2 - 3(n) + 1

Adding both sides of the above equations, we get

13 - 03 + 23 - 13 + 33 - 23 + (n-1)3 - (n-2)3 + n3 - (n - 1)3 = 3(12 + 22 + 32 + ... + n2) - 3(1+ 2 + 3 + ... + n) + n

n3  = 3(12 + 22 + 32 + ... + n2) - 3(1+ 2 + 3 + ... + n) + n

⇒ 3(12 + 22 + 32 + ... + n2) = n3 + 3(1+ 2 + 3 + ... + n) - n

⇒ 3Sn = n3 + 3(n(n-1) / 2) - n [Since 1+ 2 + 3 + ... + n = n(n+1)/2]

⇒ 3Sn = (2n3+3n2+3n-2n)/2 = (2n3+3n2+n)/2

⇒ Sn = n(2n2+3n+1)/6

⇒ Sn = n(2n2+2n+n+1)/6

⇒ Sn =   n(n+1)(2n+1) 6  

Hence, the sum of the squares of the first n natural numbers is  n(n+1)(2n+1) 6

12 + 22 + 32 + .... + n2 = ∑ k = 1 n k 2

Example: Find the sum of the squares of the first ten natural numbers.

Let S10 denote the sum of the squares of the first ten natural numbers.

Put n = 10 in Sn

S10 = 10(10+1)(2x10+1)/6 = 385

Sum of the cubes of the first n natural numbers

Sn = 13 + 23 + 33 + .... + n3

Now, we use the identity, (k+1)4 - k4 = 4k3 + 6k2 + 4k + 1 .... (i)

Putting k = 1,2,3,...,n successively in (i), we get

(2)4 - (1)4 = 4(1)3 + 6(1)2 + 4(1) + 1

(3)4 - (2)4 = 4(2)3 + 6(2)2 + 4(2) + 1

(4)4 - (3)4 = 4(3)3 + 6(3)2 + 4(3) + 1

…………………………….

…………………………….

(n + 1)4 - (n)4 = 4n3 + 6n2 + 4n + 1

Adding both sides of the above identities, we get

(n + 1)4 - (1)4 = 4(13 + 23 + 33 + .... + n3) + 6(12 + 22 + 32 + .... + n2) + 4(1 + 2 + 3 + ....+ n) + n

⇒ (n + 1)4 - (1)4 = 4Sn + 6 ∑ k = 1 n k 2 + 4 ∑ k=1 n k + n

⇒ 4Sn = (n + 1)4 - (1)4 - 6 ∑ k = 1 n k 2 - 4 ∑ k=1 n k - n
 

⇒ 4Sn = n4 + 4n3 + 6n2 + 4n - 6 x (n(n+1)(2n+1)/6) - 4 x (n(n+1)/2) - n

[ Since  ∑ k=1 n k = n(n+1)/2 and ∑ k = 1 n k 2   = n(n+1)(2n+1)/6]

⇒ 4Sn = n4 + 4n3 + 6n2 + 4n - 2n3 - 3n2 - n - 2n2 - 2n -n

⇒ 4Sn = n4 + 2n3 + n2

⇒ 4Sn = n2(n2 + 2n + 1)

⇒ 4Sn = n2(n + 1)2

⇒ Sn = n2(n + 1)2/4

⇒ Sn = (n(n + 1)/2)2
Hence, the sum of the cubes of the first n natural numbers is (n(n + 1)/2)2 .

13 + 23 + 33+....+ n3 = ∑ k = 1 n k 3

Example: Find the sum of the cubes of the first seven natural numbers.

Let S7 denotes the sum of cubes first seven natural numbers.


Then, by the formula, we have S7 = (7(7+1)/2)2 = 784

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