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Question: If the sum of m consecutive odd integers is \({{m}^{4}}\), then the first integer term is A. \({{...

If the sum of m consecutive odd integers is m4{{m}^{4}}, then the first integer term is
A. m3−m−1{{m}^{3}}-m-1
B. m3−m+1{{m}^{3}}-m+1
C. m3+m−1{{m}^{3}}+m-1
D. 2m3−m+12\dfrac{2{{m}^{3}}-m+1}{2}

Explanation

Solution

We first assume the number of terms there are before the starting number of the m consecutive odd integers’ series. We form the summation of first n odd natural numbers. We find the sum of first (n+m)\left( n+m \right) odd terms and then try to find the sum of first n odd terms. We made the relationship and solved it to find the value of the unknown.

Complete step by step answer:
Although here we are dealing with m consecutive odd integers, we need to remember that the starting number is not mentioned. So, it is given that the sum of m consecutive odd integers is m4{{m}^{4}} where the starting number is random.
We can express the odd numbers in the form of (2k−1),k∈N\left( 2k-1 \right),k\in \mathbb{N}. So, tk=(2k−1){{t}_{k}}=\left( 2k-1 \right) where tk{{t}_{k}} represents the kth{{k}^{th}} term.
Let’s assume that there are n odd terms before the starting number of the m consecutive odd integers’ series.
Now we find the summation formula for odd integers. S=∑(2k−1)S=\sum{\left( 2k-1 \right)}.
We use the formula of summation of natural numbers where ∑k=k(k+1)2\sum{k}=\dfrac{k\left( k+1 \right)}{2}.
So, S=∑(2k−1)=2∑k−∑1=k(k+1)−k=k2S=\sum{\left( 2k-1 \right)}=2\sum{k}-\sum{1}=k\left( k+1 \right)-k={{k}^{2}}.
Now we present the whole scenario where the sum of m consecutive odd integers’ series with random starting point is m4{{m}^{4}}. There are n odd terms before the starting number of the m consecutive odd integers’ series.
So, in total there will be (n+m)\left( n+m \right) terms.
We will try to find the sum of first (n+m)\left( n+m \right) odd terms and then try to find the sum of first n odd terms. We already got that the sum value will be k2,k∈N{{k}^{2}},k\in \mathbb{N}.
The sum of first (n+m)\left( n+m \right) odd terms will be (n+m)2{{\left( n+m \right)}^{2}} and the sum of first n odd terms is n2{{n}^{2}}.
We have been given a sum of m consecutive odd integers m4{{m}^{4}}.
Making the equation form we get (n+m)2−n2=m4{{\left( n+m \right)}^{2}}-{{n}^{2}}={{m}^{4}}. We solve it to get the value of n.
(n+m)2−n2=m4 ⇒(n+m−n)(n+m+n)=m4 ⇒m(2n+m)=m4 ⇒2n+m=m3 ⇒n=m3−m2 \begin{aligned} & {{\left( n+m \right)}^{2}}-{{n}^{2}}={{m}^{4}} \\\ & \Rightarrow \left( n+m-n \right)\left( n+m+n \right)={{m}^{4}} \\\ & \Rightarrow m\left( 2n+m \right)={{m}^{4}} \\\ & \Rightarrow 2n+m={{m}^{3}} \\\ & \Rightarrow n=\dfrac{{{m}^{3}}-m}{2} \\\ \end{aligned}
We found the value of n. There were n odd terms before the starting number of the m consecutive odd integers’ series.
So, the starting number of the series is the (n+1)th{{\left( n+1 \right)}^{th}} term of the odd number series.
We place the value of (n+1)\left( n+1 \right) in the formula of tk=(2k−1){{t}_{k}}=\left( 2k-1 \right).
So, (n+1)th{{\left( n+1 \right)}^{th}} term of the odd number series is tn+1=2(n+1)−1=2n+1{{t}_{n+1}}=2\left( n+1 \right)-1=2n+1.
We now use n=m3−m2n=\dfrac{{{m}^{3}}-m}{2} in tn+1=2n+1{{t}_{n+1}}=2n+1.
Therefore, tn+1=2n+1=2(m3−m2)+1=m3−m+1{{t}_{n+1}}=2n+1=2\left( \dfrac{{{m}^{3}}-m}{2} \right)+1={{m}^{3}}-m+1.

So, the correct answer is “Option B”.

Note: We can use as many variables we need but it’s more convenient to use the same variable for the summation and presenting the sequence as that helps in recognising the relation. Also, in the solution new could eliminate the value of m as m belongs to the natural number.