Question

Write the n
th term of
the series,3/7.11^2+5/8.12^2+7/9.13^2+...​

Answer 1

$\dfrac{3}{7}.11^2+\dfrac{5}{8}.12^2+\dfrac{7}{9}.13^2+......$

The given terms have 2 parts:

1. Fractional part i.e. of the type $\frac{p}{q}$.

2. Exponential part i.e. of the type $a^{b}$.

Let us consider the fractional part first:

$\dfrac{3}{7}, \dfrac{5}{8}, \dfrac{7}{9}$......

Let us further consider Numerator part only:

3, 5, 7, .....

It is an Arithmetic progression, with

First term, a = 3

Common difference, d = 2

Formula for n$_{th}$ term of an AP:

$a_n=a+(n-1)d$

So, nth term for numerator:

$\Rightarrow 3+(n-1)2\\\Rightarrow 3+2n-1\\\Rightarrow 2n+1$

Let us consider Denominator part only:

7,8,9, .....

It is an Arithmetic progression, with

First term, a = 7

Common difference, d = 1

Formula for n$_{th}$ term of an AP:

$a_n=a+(n-1)d$

So, nth term for Denominator :

$\Rightarrow 7+(n-1)1\\\Rightarrow 7+n-1\\\Rightarrow n+6$

So, nth term for fractional part:

$\dfrac{2n+1}{n+6} ...... (1)$

Now, let us consider the exponential part:

$11^{2} ,12^{2} ,13^{2} .....$

Without the square values: 11,12,13.....

It is again an Arithmetic Progression with

a = 11

d = 1

nth term:

$11+(n-1)1\\\Rightarrow 11+n-1\\\Rightarrow n+10$

nth term for exponential part: $(n+10)^{2}$ .... (2)

Using equations (1) and (2):

So, nth term for the given terms:

$\dfrac{2n+1}{n+6}. (n+10)^2$