An algebra problem by Aareyan Manzoor

Algebra Level 3

which of these are equal to 1 n 2 ( n + 1 ) + 1 n ( n + 1 ) 2 \frac{1}{n^2(n+1)}+\frac{1}{n(n+1)^2}

i) 1 n 2 ( n − 1 ) 2 + 2 n ( n − 1 ) 2 \frac{1}{n^2(n-1)^2}+\frac{2}{n(n-1)^2}
ii) 1 n 2 ( n + 1 ) 2 + 2 n ( n + 1 ) 2 \frac{1}{n^2(n+1)^2}+\frac{2}{n(n+1)^2}
iii) 1 n 2 ) − 1 n + 1 ) 2 \frac{1}{n^2)}-\frac{1}{n+1)^2}
Iv) ( 1 n + 1 + 1 n ) ( 1 n − 1 n + 1 ) (\frac {1}{n+1} +\frac{1}{n})(\frac{1}{n}-\frac{1}{n+1})

ii ,iii and iv ii,iii i,iii and iv none of the given

Aareyan Manzoor by Aareyan Manzoor , 18, USA

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1 solution

Anandhu Raj
Jan 25, 2015

Just give n any integral value,say 1 and find the value of 1 n 2 ( n + 1 ) + 1 n ( n + 1 ) 2 \frac{1}{n^2(n+1)}+\frac{1}{n(n+1)^2}

If n is given value as 1,then 1 n 2 ( n + 1 ) + 1 n ( n + 1 ) 2 \frac{1}{n^2(n+1)}+\frac{1}{n(n+1)^2} = 3 4 \frac { 3 }{ 4 }

Now similarly give options also n=1 and find out the options which give value as 3 4 \frac { 3 }{ 4 }

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