Notes On Sum of n Terms of Special Series - CBSE Class 11 Maths
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  =  $\text{{}\begin{array}{c}\text{na; if r=1}\\ \frac{\text{a(1-r}ⁿ\text{)}}{\text{1-r}}\text{; if r<1}\\ \frac{\text{a(r}ⁿ\text{-1)}}{\text{r-1}}\text{; if r>1}\end{array}\text{}}$ 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 = $\sum _{\text{k=1}}^{\text{n}}\text{k}$  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 =  $\frac{\text{n(n+1)(2n+1)}}{\text{6}}$   Hence, the sum of the squares of the first n natural numbers is $\frac{\text{n(n+1)(2n+1)}}{\text{6}}$ 12 + 22 + 32 + .... + n2 = $\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$ 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$\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$ + 4$\sum _{\text{k=1}}^{\text{n}}\text{k}$ + n ⇒ 4Sn = (n + 1)4 - (1)4 - 6$\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$ - 4$\sum _{\text{k=1}}^{\text{n}}\text{k}$ - n   ⇒ 4Sn = n4 + 4n3 + 6n2 + 4n - 6 x (n(n+1)(2n+1)/6) - 4 x (n(n+1)/2) - n [ Since $\sum _{\text{k=1}}^{\text{n}}\text{k}$ = n(n+1)/2 and $\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$  = 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 =$\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{3}}$ 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

#### 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  =  $\text{{}\begin{array}{c}\text{na; if r=1}\\ \frac{\text{a(1-r}ⁿ\text{)}}{\text{1-r}}\text{; if r<1}\\ \frac{\text{a(r}ⁿ\text{-1)}}{\text{r-1}}\text{; if r>1}\end{array}\text{}}$ 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 = $\sum _{\text{k=1}}^{\text{n}}\text{k}$  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 =  $\frac{\text{n(n+1)(2n+1)}}{\text{6}}$   Hence, the sum of the squares of the first n natural numbers is $\frac{\text{n(n+1)(2n+1)}}{\text{6}}$ 12 + 22 + 32 + .... + n2 = $\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$ 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$\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$ + 4$\sum _{\text{k=1}}^{\text{n}}\text{k}$ + n ⇒ 4Sn = (n + 1)4 - (1)4 - 6$\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$ - 4$\sum _{\text{k=1}}^{\text{n}}\text{k}$ - n   ⇒ 4Sn = n4 + 4n3 + 6n2 + 4n - 6 x (n(n+1)(2n+1)/6) - 4 x (n(n+1)/2) - n [ Since $\sum _{\text{k=1}}^{\text{n}}\text{k}$ = n(n+1)/2 and $\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{2}}$  = 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 =$\sum _{\text{k = 1}}^{\text{n}}{\text{k}}^{\text{3}}$ 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

Next