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Find the sum to `n` terms of the series: `1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+`

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To find the sum to `n` terms of the series: \[ S_n = \frac{1}{1 + 1^2 + 1^4} + \frac{2}{1 + 2^2 + 2^4} + \frac{3}{1 + 3^2 + 3^4} + \ldots + \frac{n}{1 + n^2 + n^4} \] we can follow these steps: ...
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RD SHARMA ENGLISH-SOME SPECIAL SERIES-All Questions
  1. Find the sum of all possible products of the first n natural numbers t...

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  2. Sum of n terms the series : 1^2-2^2+3^2-4^2+5^2-6^2+....

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  3. Find the sum to n terms of the series: 1/(1+1^2+1^4)+2/(1+2^2+2^4)+...

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  4. Sum the following series to n terms: 4+6+9+13+18+

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  5. Find the sum : sum(r=1)^n1/((a r+b)(a r+a+b))

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  6. Find the sum to n terms of the series: 3/(1^2 .2^2)+5/(2^2 .3^2)+7/(3^...

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  7. Sum the following series to n terms: 5+7+13+31+85+

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  8. Find the sum to n terms of the series: 1/(1. 3)+1/(3. 5)+1/(5. 7)+

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  9. Find the sum to n terms of the series: 3+15+35+63+

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  10. Find the sum to n terms of the series: 1+5+12+22+35+........

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  11. Find the sum of the series: 1. n+2.(n-1)+3.(n-2)++(n-1). 2+n .1.

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  12. Find the sum of series (3^3-2^3)+(5^3-4^3)+(7^3-6^3)+... n terms.

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  13. If Sn=sum(r=1)^n(1+2+2^2+ .......+2^r)/(2^r), then Sn is equal to (a)2...

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  14. The value of sum(r=1)^n{(2r-1)a+1/(b^r)} is equal to (a)a n^2+(b^(n-...

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  15. The sum of the series: 1/((log)2 4)+1/((log)4 4)+1/((log)8 4)++1/((log...

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  16. The sum n terms of the series 1/(sqrt(1)+sqrt(3))+1/(sqrt(3)+sqrt(5))+...

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  17. Find the sum of n terms of the series 1*2*3+2*3*4+3*4*5...........

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  18. Sum the series 3. 8+6. 11+9. 14+............... to n terms.

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  19. Find the sum of n terms of the series whose n t h term is: 2n^2-3n+5

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  20. Find the sum to n terms of the series whose nth term is n^2+2^n.

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