Question

निम्नलिखित श्रेणियों के n पदों तक योग ज्ञात कीजिएं:
$\dfrac{1^3}{1} + \dfrac{1^3 + 2^3}{1 + 3} + \dfrac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + ...$

Answer 1

Answer:

Step-by-step explanation:

माना

          $T_n=\frac{1^3+2^3+3^3+....n^3}{1+3+5+.....n} \\\\=\frac{\Σn^3}{\frac{n}{2} *[2*1+(n-1)2} \\\\=\frac{\Σn^3}{\frac{n}{2} *[2+2n-2)}\\\\=\frac{\Σn^3}{n^2} \\\\=\frac{1}{n^2} *\frac{n^2(n+1)^2}{4} \\\\=\frac{1}{4} (n^2+2n+1)\\\\S_n=\frac{1}{4} [\Σn^2+2\Σn+\Σ1]\\\\=\frac{1}{4} [\frac{n(n+1)(2n+1)}{6} +\frac{2n(n+1)}{2} +n]\\\\=\frac{n}{24} [(n+1)(2n+1)+6n(n+1)+6]\\\\=\frac{n}{24} [2n^2+n+2n+1+6n+6+6]\\\\=\frac{n}{24} (2n^2+9n+13)$

Answer 2

(n/24) ( 2n² + 9n + 13)  

Step-by-step explanation:

1³/1   + (1³ + 2³) /( 1+3)  +  (1³ + 2³ + 3³) /( 1+3 + 5)

aₙ  =  (1³ + 2³ + 3³ +....................+n³)/(1 + 3 + 5  + .................+ (2n-1)

1³ + 2³ + 3³ +....................+n³ = ∑ n³ =( n(n + 1)/2)²

(1 + 3 + 5  + .................+ (2n-1)  =  (n/2)(1 + 2n - 1)  = n²

aₙ = ( n(n + 1)/2)² /  n²

aₙ = (n + 1)²/4

aₙ = (n²  + 2n + 1)/4

∑aₙ = ∑(n²  + 2n + 1)/4

= (1/4) (∑n²  + 2 ∑ n  + ∑ 1  )

=(1/4)  (  n(n+1)(2n + 1)/6  +  2n(n+1)/2  + n)

=(1/4)  (  n(n+1)(2n + 1)/6  +  n(n+1)  + n)

= (n/24) ( (n+1)(2n + 1) + 6(n+1)  + 6)

=  (n/24) ( 2n² + 3n + 1 + 6n+6  + 6)

= (n/24) ( 2n² + 9n + 13)

1³/1   + (1³ + 2³) /( 1+3)  +  (1³ + 2³ + 3³) /( 1+3 + 5)  = (n/24) ( 2n² + 9n + 13)

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