Solveeit Logo
Login

Question

Question: If \(1+2+3+........+n=k\) then \({{1}^{3}}+{{2}^{3}}+.......{{n}^{3}}\)? A) \({{k}^{2}}\) B) \(...

If 1+2+3+........+n=k1+2+3+........+n=k then 13+23+.......n3{{1}^{3}}+{{2}^{3}}+.......{{n}^{3}}?
A) k2{{k}^{2}}
B) k3{{k}^{3}}
C) k(k+1)2\dfrac{k\left( k+1 \right)}{2}
D) (k+1)3{{\left( k+1 \right)}^{3}}

Explanation

Solution

The above given question is a series. Actually the series is a sum of a sequence. Now, what is sequence? In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. For example,\left\\{ 1,2,3........ \right\\} is an example of sequence or it is an infinite sequence. Sum of the sequence is known as a series.

Complete step by step solution:
The given series is:
⇒1+2+3+........+n=k\Rightarrow 1+2+3+........+n=k
It means Sn=k{{S}_{n}}=k , the above sequence is the sum of natural numbers.
Here we have to find out the sum of the cubes of natural numbers upto n.
We know the direct formula of the sum of the natural number is:
⇒Sn=n(n+1)2\Rightarrow {{S}_{n}}=\dfrac{n\left( n+1 \right)}{2}
It is given that Sn=k{{S}_{n}}=k, so we will replace Sn{{S}_{n}} by kk in the above expression, then we get
⇒k=n(n+1)2.........(1)\Rightarrow k=\dfrac{n\left( n+1 \right)}{2}.........(1)
We know that the sum of cubes of the n natural numbers is:
⇒Sn=[n(n+1)2]2\Rightarrow {{S}_{n}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}
The given series of the sum of cubes of the n natural numbers is:
⇒Sn=13+23+.......n3\Rightarrow {{S}_{n}}={{1}^{3}}+{{2}^{3}}+.......{{n}^{3}}
Now in the place of Sn{{S}_{n}}we will put the formula of sum of cubes of n natural number, then we get
⇒Sn=13+23+........+n3=[n(n+1)2]2\Rightarrow {{S}_{n}}={{1}^{3}}+{{2}^{3}}+........+{{n}^{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}
Now from equation (1) we can easily see that k=n(n+1)2k=\dfrac{n\left( n+1 \right)}{2}, now put this value in the above equation , then we get
⇒13+23+.....+n3=[n(n+1)2]2=k2\Rightarrow {{1}^{3}}+{{2}^{3}}+.....+{{n}^{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}={{k}^{2}}
Hence, we get that the sum of cubes of the n natural number is k2{{k}^{2}}.

So, the correct answer is “Option A”.

Note: If we roughly speak the series is a description of the operation of adding infinitely many quantities, one after the other to a given starting quantity. Series are used in most areas of mathematics. Sometimes we also say series is a partial sum of infinite series.