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
= $\sum\limits_{k\,=\,1}^{k\,=\,n}{k^3}~=~\left[ \frac{n (n + 1)}{2}\right]^2$
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
⇒ an = 5 + (n-1)(4+n)
⇒ an = 5 + 4n + n2 - 4 - n
⇒ an = 1 + 3n + n2
⇒ an = n2 + 3n + 1
7. Thus we obtained the nth term.
• nth term = an = n2 + 3n + 1.
• So the sum (Sn) 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 nth term is n(n+3)
Solution:
• nth term = n(n+3).
• So the sum (Sn) 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.
The link below gives some more solved examples
In the next section we will see some miscellaneous examples.
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