Question

Find the sum of 'n' twrms of the series :-


$\frac{2 {a}^{2} - 1 }{a} , \: (4a - \frac{3}{a}) , \: \frac{6 {a}^{2} - 5}{a} , \: .................$



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Answer 1

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Answer 2

Answer:

$\mathsf{\dfrac{2a^2-1}{a}+4a-\dfrac{3}{a}+\dfrac{6a^2-5}{a}+.......n\:times}\\\\\mathsf{\implies \dfrac{2a^2-1}{a}+\dfrac{4a^2-3}{a}+...n\:times}\\\\\textbf{Comparing this with an AP we can write the following :}\\\\\bigstar \mathtt{First\:term(a)=\dfrac{2a^2-1}{a}}\\\\\bigstar \mathtt{Common\:difference(d)=\dfrac{2a^2-2}{a}}$

$\textbf{Sum of n terms of an AP is given by}\\\\\bigstar \mathfrak{S_n=\dfrac{n}{2}[2a+(n-1)d]}]\\\\\mathsf{\implies \dfrac{n}{2}[2(\dfrac{2a^2-1}{a})+(n-1)(\dfrac{2a^2-2}{a})]}\\\\\implies \mathsf{\dfrac{n}{2a}[4a^2-2+2(a^2-1)(n-1)]}\\\\\implies \mathsf{\dfrac{n}{a}[2a^2-1+(a^2-1)(n-1)]}\\\\\implies \mathsf{\dfrac{n}{a}[a^2+a^2-1+(a^2-1)(n-1)]}\\\\\implies \mathsf{\dfrac{n}{a}[a^2+(a^2-1)(n-1+1)]}\\\\\implies \mathsf{\dfrac{n}{a}[a^2+n(a^2-1)]}$

$\implies \mathsf{\dfrac{n}{a}[a^2+a^2n-n]}\\\\\implies \mathsf{\dfrac{n}{a}[a^2(n+1)-n]}\\\\\implies \boxed{\mathtt{na(n+1)-\dfrac{n^2}{a}}}$

Step-by-step explanation:

We are given a series and we are asked to find the Sum of n terms of the given series .

Use the formula for the Sum of n terms and then we simplify the problem further and ultimately get the required answer !