Answer

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**Hint:**We solve this problem first by finding the \[{{n}^{th}}\] term. Then we use the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is

\[{{S}_{n}}=\sum{{{T}_{n}}}\]

We use some standard results of sum of terms that is

\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]

\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]

\[\sum{1}=n\]

By using the above results we find the sum of terms to get the required value.

**Complete step by step answer:**

We are given that the series as \[4+6+9+13+18+..........\]up to \['n'\] terms

Let us assume that the sum of the given series as

\[\Rightarrow {{S}_{n}}=4+6+9+13+.............+{{T}_{n-1}}+{{T}_{n}}...........equation(i)\]

Now let us add ‘0’ on both sides so that the result will not change that is

\[\Rightarrow {{S}_{n}}=0+4+6+9+13+............+{{T}_{n-1}}+{{T}_{n}}...........equation(ii)\]

By subtracting the equation (ii) from equation (i) we get

\[\begin{align}

& \Rightarrow 0=4+\left( 2+3+4+...........\left( n-1 \right)terms \right)-{{T}_{n}} \\

& \Rightarrow {{T}_{n}}-4=2+3+4+...........\left( n-1 \right)terms \\

\end{align}\]

Here we can see that the RHS is in A.P

We know that the sum of \['n'\] terms in an A.P having \['a'\] as first term and \['d'\] as common difference is given as

\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]

By using this formula to above equation we get

\[\begin{align}

& \Rightarrow {{T}_{n}}-4=\dfrac{n-1}{2}\left( 2\times 2+\left( n-1-1 \right)\times 1 \right) \\

& \Rightarrow {{T}_{n}}-4=\dfrac{\left( n-1 \right)\left( n+2 \right)}{2} \\

\end{align}\]

Now, by multiplying the terms on RHS we get

\[\begin{align}

& \Rightarrow {{T}_{n}}=\dfrac{1}{2}\left( {{n}^{2}}+n-2 \right)+4 \\

& \Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{2}+\dfrac{n}{2}+3 \\

\end{align}\]

We know that the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is

\[{{S}_{n}}=\sum{{{T}_{n}}}\]

By using the above formula we get the sum of given series as

\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \sum{{{n}^{2}}}+\sum{n} \right)+3\sum{1}.......equation(iii)\]

We know that the some of the standard results of sum of terms that is

\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]

\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]

\[\sum{1}=n\]

By using these formulas in equation (iii) we get

\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2} \right)+3n\]

Now by taking the common terms out we get

\[\begin{align}

& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{\left( n+1 \right)}{2} \right)+3n \\

& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1}{6}+\dfrac{\left( n+1 \right)}{2}+6 \right) \\

\end{align}\]

Now, by using the LCM method and adding the terms we get

\[\begin{align}

& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1+3n+3+36}{6} \right) \\

& \Rightarrow {{S}_{n}}=\dfrac{n}{2\times 6}\left( 2{{n}^{2}}+6n+40 \right) \\

& \Rightarrow {{S}_{n}}=\dfrac{n}{6}\left( {{n}^{2}}+3n+20 \right) \\

\end{align}\]

We are given that the sum of series as

\[\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)\]

By comparing this given formula with the result we get

\[\begin{align}

& \Rightarrow m=20 \\

& \Rightarrow k=6 \\

\end{align}\]

Now by finding the value of \[m-k\] we get

\[\Rightarrow m-k=20-6=14\]

**Therefore, the value of \[m-k\] is 14.**

**Note:**We can solve this problem in other methods also.

We have the series as

\[4+6+9+13+18+..........\]

Here, we can see that the differences of consecutive terms are in A.P then the \[{{n}^{th}}\] term is given as

\[\Rightarrow {{T}_{n}}=a{{n}^{2}}+bn+c\]

By taking \[n=1\] we get

\[\Rightarrow a+b+c=4......equation(i)\]

By taking \[n=2\] we get

\[\Rightarrow 4a+2b+c=6......equation(ii)\]

By taking \[n=3\] we get

\[\Rightarrow 9a+3b+c=9......equation(iii)\]

Now by subtracting equation (i) from equation (ii) we get

\[\begin{align}

& \Rightarrow 3a+b=2 \\

& \Rightarrow b=2-3a \\

\end{align}\]

Now, by subtracting equation (ii) from equation (iii) we get

\[\begin{align}

& \Rightarrow 5a+b=3 \\

& \Rightarrow 5a+2-3a=3 \\

& \Rightarrow a=\dfrac{1}{2} \\

\end{align}\]

By substituting the value of \['a'\] in \['b'\] we get

\[\begin{align}

& \Rightarrow b=2-\dfrac{3}{2} \\

& \Rightarrow b=\dfrac{1}{2} \\

\end{align}\]

Now, from equation (i) we get

\[\begin{align}

& \Rightarrow \dfrac{1}{2}+\dfrac{1}{2}+c=4 \\

& \Rightarrow c=3 \\

\end{align}\]

Therefore the \[{{n}^{th}}\] term is given as

\[\Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{2}+\dfrac{n}{2}+3\]

We know that the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is

\[{{S}_{n}}=\sum{{{T}_{n}}}\]

By using the above formula we get the sum of given series as

\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \sum{{{n}^{2}}}+\sum{n} \right)+3\sum{1}.......equation(iii)\]

We know that the some of the standard results of sum of terms that is

\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]

\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]

\[\sum{1}=n\]

By using these formulas in equation (iii) we get

\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2} \right)+3n\]

Now by taking the common terms out we get

\[\begin{align}

& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{\left( n+1 \right)}{2} \right)+3 \\

& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1}{6}+\dfrac{\left( n+1 \right)}{2}+6 \right) \\

\end{align}\]

Now, by using the LCM method and adding the terms we get

\[\begin{align}

& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1+3n+3+36}{6} \right) \\

& \Rightarrow {{S}_{n}}=\dfrac{n}{2\times 6}\left( 2{{n}^{2}}+6n+40 \right) \\

& \Rightarrow {{S}_{n}}=\dfrac{n}{6}\left( {{n}^{2}}+3n+20 \right) \\

\end{align}\]

We are given that the sum of series as

\[\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)\]

By comparing this given formula with the result we get

\[\begin{align}

& \Rightarrow m=20 \\

& \Rightarrow k=6 \\

\end{align}\]

Now by finding the value of \[m-k\] we get

\[\Rightarrow m-k=20-6=14\]

Therefore, the value of \[m-k\] is 14.

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