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# Find the sum of (1^4/1*3)+(2^4/3*5)+(3^4/5*7)+..................+(n^4/(2n-1)(2n+1)).

Sathya
35 Points
9 years ago

If it continues up to m terms,thn

$\dpi{100} \fn_cm m^2+m/6$ is the ans i think

Because its

$\dpi{100} \fn_cm (n^4/(2n-1)(2n+1))$ +

$\dpi{100} \fn_cm ((2n)^4/((2(2n))-1)((2(2n)+1))$ and so on........

If u take $\dpi{100} \fn_cm (n^4/(2n-1)(2n+1))$  common,thn it becomes

 $\dpi{100} \fn_cm (n^4/(2n-1)(2n+1)) (1+2+3+4......M terms)$   Thn Its ans is $\dpi{100} \fn_cm (m^2+m)/6$

Swapnil Saxena
102 Points
9 years ago

@ I cant understand ur solution. Pls simplify it...

The solution fr the ques is here it is n[{2(n+1)(2n+1)}/3 + 1/(2n+1) +1]/16

Sathya
35 Points
9 years ago

$\LARGE \dpi{50} (m^4 + 2m^3 +2m^2+m)/6(2m+1)$