Question

Sum of the series $4+6+9+13+18+..........$up to $'n'$ terms be $\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)$ then find the value of $m-k$

Answer 1

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

& \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{aligned}$$ Here we can see that the RHS is in A.P We know that the sum of $$'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{aligned} & \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{aligned}$$ Now, by multiplying the terms on RHS we get $$\begin{aligned} & \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{aligned}$$ We know that the standard formula of sum of n terms using the $${{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{aligned} & \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{aligned}$$ Now, by using the LCM method and adding the terms we get $$\begin{aligned} & \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{aligned}$$ We are given that the sum of series as $$\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)$$ By comparing this given formula with the result we get $$\begin{aligned} & \Rightarrow m=20 \\\ & \Rightarrow k=6 \\\ \end{aligned}$$ Now by finding the value of $$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{aligned} & \Rightarrow 3a+b=2 \\\ & \Rightarrow b=2-3a \\\ \end{aligned}$$ Now, by subtracting equation (ii) from equation (iii) we get $$\begin{aligned} & \Rightarrow 5a+b=3 \\\ & \Rightarrow 5a+2-3a=3 \\\ & \Rightarrow a=\dfrac{1}{2} \\\ \end{aligned}$$ By substituting the value of $$'a'$$ in $$'b'$$ we get $$\begin{aligned} & \Rightarrow b=2-\dfrac{3}{2} \\\ & \Rightarrow b=\dfrac{1}{2} \\\ \end{aligned}$$ Now, from equation (i) we get $$\begin{aligned} & \Rightarrow \dfrac{1}{2}+\dfrac{1}{2}+c=4 \\\ & \Rightarrow c=3 \\\ \end{aligned}$$ Therefore the $${{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{aligned} & \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{aligned}$$ Now, by using the LCM method and adding the terms we get $$\begin{aligned} & \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{aligned}$$ We are given that the sum of series as $$\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)$$ By comparing this given formula with the result we get $$\begin{aligned} & \Rightarrow m=20 \\\ & \Rightarrow k=6 \\\ \end{aligned}$$ Now by finding the value of $$m-k$$ we get $$\Rightarrow m-k=20-6=14$$ Therefore, the value of $$m-k$$ is 14.