Chapter 9.6 - Sum of The Cubes of First n Natural Numbers

In the previous section, we saw the sum of squares of the first n natural numbers. In this section, we will see sum of cubes of the first n natural numbers. 

C. 1³ + 2³ + 3³ +. . . + n³

This sum can be calculated in 8 steps:
1. Consider the identity: (a+b)⁴ = a⁴+4a³b+6a²b²+4ab³+b⁴
[Recall that (a+b) can be raised to any power by using binomial theorem that we saw in the previous chapter]
Let us put a = k and b = -1. We get:

$\begin{array}{ll} {}&{[k+(-1)]^4} &{}={}& {[k-1]^4} &{}& {} &{}& {} &{}& {}&{}& {} &{} \\ {}&{} &{}={}& {k^4 ~+~ 4 × k^3 × -1 ~+~6 × k^2 × (-1)^2~+~4 × k × (-1)^3~+~(-1)^4} &{}& {} &{}{}& {} &{}& {}&{}& {} &{} \\ {}&{} &{}={}& {k^4 ~-~ 4 k^3~+~6k^2 ~-~4k~+~1} &{}& {} &{}{}& {} &{}& {}&{}& {} &{} \\ \end{array}$

2. Subtracting (k-1)⁴ from k⁴, we get:

$\begin{array}{ll} {}&{k^4~-~(k-1)^4} &{}={}& {k^4~-~\left(k^4 ~-~ 4 k^3~+~6k^2 ~-~4k~+~1 \right)} &{}& {} &{}& {} &{}& {}&{}& {} &{} \\ {}&{} &{}={}& {k^4~-~k^4~+~4 k^3~-~6k^2 ~+~4k~-~1} &{}& {} &{}{}& {} &{}& {}&{}& {} &{} \\ {}&{} &{}={}& {4 k^3~-~6k^2 ~+~4k~-~1} &{}& {} &{}{}& {} &{}& {}&{}& {} &{} \\ \end{array}$

• We can use this as an identity:
$k^4~-~(k-1)^4~=~4 k^3~-~6k^2 ~+~4k~-~1$

3. In the above identity, let us put k = 1, 2, 3, . . . , n successively. We get:

$\begin{array}{ll} {\text{When k = 1,}}&{1^4~-~(1-1)^4} &{}={}& {1^4~-~0^4} &{}={}& {4 × 1^3~-~6 × 1^2 ~+~4 × 1~-~1} &{}={}& {4(1)^3~-~6(1)^2~+~4(1)~-~1} &{}& {}&{}& {} &{} \\ {\text{When k = 2,}}&{2^4~-~(2-1)^4} &{}={}& {2^4~-~1^4} &{}={}& {4 × 2^3~-~6 × 2^2 ~+~4 × 2~-~1} &{}={}& {4(2)^3~-~6(2)^2~+~4(2)~-~1} &{}& {}&{}& {} &{} \\ {\text{When k = 3,}}&{3^4~-~(3-1)^4} &{}={}& {3^4~-~2^4} &{}={}& {4 × 3^3~-~6 × 3^2 ~+~4 × 3~-~1} &{}={}& {4(3)^3~-~6(3)^2~+~4(3)~-~1} &{}& {}&{}& {} &{} \\ {\text{When k = 4,}}&{4^4~-~(4-1)^4} &{}={}& {4^4~-~3^4} &{}={}& {4 × 4^3~-~6 × 4^2 ~+~4 × 4~-~1} &{}={}& {4(4)^3~-~6(4)^2~+~4(4)~-~1} &{}& {}&{}& {} &{} \\ {-}&{-} &{}& {-} &{}& {-} &{}& {-} &{}& {}&{}& {} &{} \\ {-}&{-} &{}& {-} &{}& {-} &{}& {-} &{}& {}&{}& {} &{} \\ {\text{When k = n,}}&{n^4~-~(n-1)^4} &{}& {} &{}={}& {4 × n^3~-~6 × n^2 ~+~4 × n~-~1} &{}={}& {4(n)^3~-~6(n)^2~+~4(n)~-~1} &{}& {}&{}& {} &{} \\ \end{array}$

4. Picking the first and last items from each line, we get:

$\begin{array}{ll} {}&{} &{}& {1^4~-~0^4} &{}& {} &{}={}& {4(1)^3~-~6(1)^2~+~4(1)~-~1} &{}& {}&{}& {} &{} \\ {}&{} &{}& {2^4~-~1^4} &{}& {} &{}={}& {4(2)^3~-~6(2)^2~+~4(2)~-~1} &{}& {}&{}& {} &{} \\ {}&{} &{}& {3^4~-~2^4} &{}& {} &{}={}& {4(3)^3~-~6(3)^2~+~4(3)~-~1} &{}& {}&{}& {} &{} \\ {}&{} &{}& {4^4~-~3^4} &{}& {} &{}={}& {4(4)^3~-~6(4)^2~+~4(4)~-~1} &{}& {}&{}& {} &{} \\ {}&{} &{}& {-} &{}& {} &{}& {-} &{}& {}&{}& {} &{} \\ {}&{} &{}& {-} &{}& {} &{}& {-} &{}& {}&{}& {} &{} \\ {}&{} &{}& {n^4~-~(n-1)^4} &{}& {} &{}={}& {4(n)^3~-~6(n)^2~+~4(n)~-~1} &{}& {}&{}& {} &{} \\ \end{array}$

5. In the above result in (4), let us add all terms on the left side.
• We see that diagonal elements will get cancelled:
   ♦ 1⁴ will get cancelled by -1⁴.
   ♦ 2⁴ will get cancelled by -2⁴.
   ♦ 3⁴ will get cancelled by -3⁴.
   ♦ so on . . .
• So only 0⁴ and n⁴ will remain.
• Thus the sum of all terms on the left side is: (n⁴ - 0⁴) = n⁴.

6. In the result in (4), let us add all terms on the right side. We get:
4(1³ + 2³ + 3³ +. . . + n³) - 6(1² + 2² + 3² +. . . + n²) + 4(1 + 2 + 3 +. . . + n) - (1+1+1+ . . . n times)
• This can be written in a shortened form using sigma notations:
$4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}~-~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n$

7. Equating the results in (5) and (6), we get:

$\begin{array}{ll} {}&{n^4} &{}={}& {4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}~-~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n} &{}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^4~+~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~-~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n} &{\color {green} {\text{- - - - (a)}}}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^4~+~\left[\frac{6n(2n + 1) (n+1)}{6} \right]~-~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n} &{\color {green} {\text{- - - - (b)}}}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^4~+~\left[\frac{6n(2n + 1) (n+1)}{6} \right]~-~\left[\frac{4n(n+1)}{2} \right]~+~n} &{}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^4~+~n(2n + 1) (n+1)~-~2n(n+1)~+~n} &{}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^4~+~2n^3+2n^2+n^2+n~-~2n^2 -2n~+~n} &{}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^4 + 2n^3 + n^2 } &{}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^2 \left(n^2 + 2n + 1 \right) } &{\color {green} {\text{- - - - (c)}}}& {}  \\ {\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {n^2 \left(n + 1 \right)^2 } &{}& {}  \\ {\Rightarrow}&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {\frac{n^2 \left(n + 1 \right)^2}{4} } &{}& {}  \\ {\Rightarrow}&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}} &{}={}& {\left[ \frac{n (n + 1)}{2}\right]^2} &{}& {}  \\ \end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• Using the identity (a+b)² = a² + 2ab + b²,
(n² + 2n + 1) is (n+1)²

8. We can write the result as a formula:
Sum of cubes of the first n natural numbers
= 1³ + 2³ + 3³ +. . . + n³

= $\sum\limits_{k\,=\,1}^{k\,=\,n}{k^3}~=~\left[ \frac{n (n + 1)}{2}\right]^2$

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Let us see some solved examples:

Solved example 9.19
Find the sum to n terms of the series: 5 + 11 + 19 + 29 + 41 + . . . 
Solution:
1. Let us write:
$S_n~=~5~+~11~+~19~+~29~+~41~+~.~.~.~+~a_{n-1}~+~a_{n}$
2. The same result can be written as:
$S_n~=~5~+~11~+~19~+~29~+~41~+~.~.~.~+~a_{n-2}~+~a_{n-1}~+~a_{n}$
3. We will write the second result just below the first result. But one term to the right. We get:

$\begin{array}{ll} {S_n}&{}={} &{5}& {~+~} &{11}& {~+~} &{19}& {~+~} &{29}& {~+~.~.~.~+~}&{a_{n-1}}& {~+~} &{a_n} &{} &{} \\ {S_n}&{}={} &{}& {} &{5}& {~+~} &{11}& {~+~} &{19}& {~+~.~.~.~+~}&{a_{n-2}}& {~+~} &{a_{n-1}} &{~+~} &{a_n} \\ \end{array}$

4. Subtracting each term in the second row, from the term directly above it, we get:

$\begin{array}{ll} {S_n}&{}={} &{5}& {~+~} &{11}& {~+~} &{19}& {~+~} &{29}& {~+~.~.~.~+~}&{a_{n-1}}& {~+~} &{a_n} &{} &{} \\ {S_n}&{}={} &{}& {} &{5}& {~+~} &{11}& {~+~} &{19}& {~+~.~.~.~+~}&{a_{n-2}}& {~+~} &{a_{n-1}} &{~+~} &{a_n} \\ {0}&{}={} &{5}& {~+~} &{\left\{6 \right. }& {~+~} &{8}& {~+~} &{10}& {~+~.~.~.~}&{}& {} &{\left. \right \}} &{~-~} &{a_n} \\ \end{array}$

5. In the above result, there will be (n-1) terms inside the curly brackets '{}'
• In side the curly brackets, we have the series: 6 + 8 + 10 + . . . (n-1) terms
• It is an A.P with a = 6, d = 2 and n = (n-1)
• So sum of the series = $\frac{(n-1)}{2} [2 × 6 + (n-1-1)2]~=~\frac{(n-1)}{2} [12 + (n-2)2]~=~(n-1)(6+n-2)~=~(n-1)(4+n)$

6. Now the result in (4) becomes:
0 = 5 + (n-1)(4+n) - an
⇒ a_n = 5 + (n-1)(4+n)
⇒ a_n = 5 + 4n + n² - 4 - n
⇒ a_n = 1 + 3n + n²
⇒ a_n = n² + 3n + 1

7. Thus we obtained the n^th term.
• n^th term = a_n = n² + 3n + 1.
• So the sum (S_n) of the given series can be obtained as:

$\begin{array}{ll} {S_n}&{}={} &{\sum\limits_{k\,=\,0}^{k\,=\,n}{a_k}}~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{\left(k^2 + 3k + 1 \right)}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}& {} &{\color {green} {\text{- - - - (a)}}}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n(2n + 1) (n+1)}{6}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}& {} &{\color {green} {\text{- - - - (b)}}}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n(2n + 1) (n+1)}{6}~+~ \frac{3n (n+1)}{2}~+~n}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n(2n + 1) (n+1)~+~9n(n+1)~+~6n}{6}}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n[(2n+1)(n+1)+9(n+1)+6]}{6}}& {}={} &{\frac{n[2n^2+2n+n+1+9n+9+6]}{6}}& {}={} &{\frac{n[2n^2+12n+16]}{6}}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{2n[n^2+6n+8]}{6}}& {}={} &{\frac{n[n^2+6n+8]}{3}}& {} &{\color {green} {\text{- - - - (c)}}}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n[(n+2)(n+4)]}{3}}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ \end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• n² + 6n + 8 = 0 can be solved as a quadratic equation.
• We will get: n = -2 and n = -4.

Solved example 9.20
Find the sum to n terms of the series whose n^th term is n(n+3)
Solution:
• n^th term = n(n+3).
• So the sum (S_n) of the given series can be obtained as:

$\begin{array}{ll} {S_n}&{}={} &{\sum\limits_{k\,=\,0}^{k\,=\,n}{[k(k+3)]}}~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{\left[k^2 + 3k \right]}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}}& {} &{\color {green} {\text{- - - - (a)}}}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n(2n + 1) (n+1)}{6}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}}& {} &{\color {green} {\text{- - - - (b)}}}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n(2n + 1) (n+1)}{6}~+~ \frac{3n (n+1)}{2}}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n(2n + 1) (n+1)~+~9n(n+1)}{6}}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n[(2n+1)(n+1)+9(n+1)]}{6}}& {}={} &{\frac{n[2n^2+2n+n+1+9n+9]}{6}}& {}={} &{\frac{n[2n^2+12n+10]}{6}}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{2n[n^2+6n+5]}{6}}& {}={} &{\frac{n[n^2+6n+5]}{3}}& {} &{\color {green} {\text{- - - - (c)}}}& {} &{}& {}&{}& {} &{} &{} &{} \\ {}&{}={} &{\frac{n[(n+1)(n+5)]}{3}}& {} &{}& {} &{}& {} &{}& {}&{}& {} &{} &{} &{} \\ \end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• n² + 6n + 5 = 0 can be solved as a quadratic equation.
• We will get: n = -1 and n = -5.

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The link below gives some more solved examples

Exercise 9.4

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In the next section we will see some miscellaneous examples.

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