Question

Let the greatest common divisor of m,n be 1 if 1 /1.7 + 1 /7.13 + 1 /13.19 +.... upto 20 terms = m/n then 5m+ 2n =​

Answer 1

Given,

$\small\longrightarrow\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+\,\dots\,20\ terms=\dfrac{m}{n}$

Let us find the general term of the series first.

The sequence 1, 7, 13, 19,... is an AP of first term 1 and common difference 6. So its general term is 6n + (1 - 6) = 6n - 5.

Then the general term of our series should be,

$\small\displaystyle\longrightarrow\sum_{n=1}^{20}\dfrac{1}{(6n-5)(6(n+1)-5)}=\dfrac{m}{n}$

$\small\displaystyle\longrightarrow\sum_{n=1}^{20}\dfrac{1}{(6n-5)(6n+1)}=\dfrac{m}{n}$

We see that (6n + 1) - (6n - 5) = 6. So one can write,

$\small\displaystyle\longrightarrow\dfrac{1}{6}\sum_{n=1}^{20}\dfrac{(6n+1)-(6n-5)}{(6n-5)(6n+1)}=\dfrac{m}{n}$

$\small\displaystyle\longrightarrow\dfrac{1}{6}\sum_{n=1}^{20}\left(\dfrac{1}{6n-5}-\dfrac{1}{6n+1}\right)=\dfrac{m}{n}$

$\small\displaystyle\longrightarrow\dfrac{1}{6}\left(\sum_{n=1}^{20}\dfrac{1}{6n-5}-\sum_{n=1}^{20}\dfrac{1}{6n+1}\right)=\dfrac{m}{n}$

$\small\displaystyle\longrightarrow\dfrac{1}{6}\left(\sum_{n=1}^{20}\dfrac{1}{6n-5}-\sum_{n=1}^{20}\dfrac{1}{6(n+1)-5}\right)=\dfrac{m}{n}$

We have,

• $\small\displaystyle\sum_{n=a}^bf(n)=\sum_{n=a+k}^{b+k}f(n-k)$

According to this formula, we replace n in the second sum by n-1 so the limit gets added by 1 also.

$\small\displaystyle\longrightarrow\dfrac{1}{6}\left(\sum_{n=1}^{20}\dfrac{1}{6n-5}-\sum_{n=1+1}^{20+1}\dfrac{1}{6(n-1+1)-5}\right)=\dfrac{m}{n}$

$\small\displaystyle\longrightarrow\dfrac{1}{6}\left(\sum_{n=1}^{20}\dfrac{1}{6n-5}-\sum_{n=2}^{21}\dfrac{1}{6n-5}\right)=\dfrac{m}{n}$

We take the term having n = 1 out of the first sum and the term having n = 21 out of the second sum.

$\small\displaystyle\longrightarrow\dfrac{1}{6}\left(\dfrac{1}{6(1)-5}+\sum_{n=2}^{20}\dfrac{1}{6n-5}-\sum_{n=2}^{20}\dfrac{1}{6n-5}-\dfrac{1}{6(21)-5}\right)=\dfrac{m}{n}$

$\small\displaystyle\longrightarrow\dfrac{1}{6}\left(1-\dfrac{1}{121}\right)=\dfrac{m}{n}$

$\small\displaystyle\longrightarrow\dfrac{20}{121}=\dfrac{m}{n}$

Since gcd(20, 121) = 1,

• $\smallm=20$

• $\smalln=121$

Hence,

$\small\displaystyle\longrightarrow 5m+2n=5(20)+2(121)$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{5m+2n=342}} }$

Answer 2

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