Question

If it is given that $\left( {{1}^{2}}-{{t}_{1}} \right)+\left( {{2}^{2}}-{{t}_{2}} \right)+...+\left( {{n}^{2}}-{{t}_{n}} \right)=\dfrac{n\left( {{n}^{2}}-1 \right)}{3}$, then what is the value of ${{t}_{n}}$.
(a) ${{n}^{2}}$
(b) $2n$
(c) ${{n}^{2}}$ - $2n$
(d) none of these

Answer 1

Hint: To solve for this problem we will split this problem into two series one in which squares of each term is added and other in which terms ${{t}_{1}},{{t}_{2}}...{{t}_{n}}$ are added. Now we will assume ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}+...+{{t}_{n}}$ equal to ${{S}_{n}}$ and we will solve the both sides of equations to find the value of ${{S}_{n}}$. Then the term ${{t}_{n}}$ will be equal to ${{S}_{n}}-{{S}_{n-1}}$.

Complete step by step answer:
We are given the expression,
$\left( {{1}^{2}}-{{t}_{1}} \right)+\left( {{2}^{2}}-{{t}_{2}} \right)+...+\left( {{n}^{2}}-{{t}_{n}} \right)=\dfrac{n\left( {{n}^{2}}-1 \right)}{3}$ and
We have to find the value of ${{t}_{n}}$,
First we will split the LHS of the equation into two parts where first part will be the sum of the squares of natural numbers and second part will be the sum of terms ${{t}_{1}},{{t}_{2}}...{{t}_{n}}$.
So by arranging the terms, we get
$\left( {{1}^{2}}+{{2}^{2}}+...+{{n}^{2}} \right)-\left( {{t}_{1}}+{{t}_{2}}+...+{{t}_{n}} \right)=\dfrac{n\left( {{n}^{2}}-1 \right)}{3}$
We know sum of squares of first n natural numbers is given by $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ so putting this in equation, we get
\[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-\left( {{t}_{1}}+{{t}_{2}}+...+{{t}_{n}} \right)=\dfrac{n\left( {{n}^{2}}-1 \right)}{3}\]
Now we will suppose ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}+...+{{t}_{n}}$ equal to ${{S}_{n}}$, so we get
\[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-{{S}_{n}}=\dfrac{n\left( {{n}^{2}}-1 \right)}{3}\]
${{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{n\left( {{n}^{2}}-1 \right)}{3}$
We will further solve this expression as,
\[\begin{align}
  & {{S}_{n}}=\dfrac{2{{n}^{3}}+3{{n}^{2}}+n}{6}-\dfrac{{{n}^{3}}-n}{3} \\
 & {{S}_{n}}=\dfrac{2{{n}^{3}}+3{{n}^{2}}+n-2{{n}^{3}}+2n}{6} \\
 & {{S}_{n}}=\dfrac{3{{n}^{2}}+3n}{6} \\
 & {{S}_{n}}=\dfrac{{{n}^{2}}+n}{2} \\
\end{align}\]
${{S}_{n}}=\dfrac{n\left( n+1 \right)}{2}$
Now ${{n}^{th}}$ term of a series is given by ${{S}_{n}}-{{S}_{n-1}}$, so we get
${{t}_{n}}={{S}_{n}}-{{S}_{n-1}}=\dfrac{n\left( n+1 \right)}{2}-\dfrac{n\left( n-1 \right)}{2}$
Solving the above expression as following we get,
\[\begin{align}
  & {{t}_{n}}=\dfrac{n\left( n+1 \right)-n\left( n-1 \right)}{2} \\
 & {{t}_{n}}=\dfrac{{{n}^{2}}+n-{{n}^{2}}+n}{2} \\
 & {{t}_{n}}=\dfrac{2n}{2} \\
\end{align}\]
${{t}_{n}}=n$
None of the given options matches our answer so the correct option is (d) i.e. none of these.

Note:
To solve this problem you need to first observe the question carefully in order to see how we can manipulate the given equations to find the answer. And you need to remember that sum of squares of first n natural numbers is given by $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ and ${{n}^{th}}$ term of a series is given by ${{S}_{n}}-{{S}_{n-1}}$.