Question

दर्शाइए कि $\dfrac{1 \times 2^2 + 2\times 3^2 + ... + n(n + 1)^2 }{1^2 \times 2 + 2^2 \times 3 + ... + n^2(n + 1)} = \dfrac{3n + 5}{3n + 1}$

Answer 1

Answer:

Step-by-step explanation:

अंश का nवां पद = n (n+1)² = n(n²+2n+1)

= n³+2n²+n

अंश के इन पदों का योग s₁ = Σn³+2Σn² +Σn

n²(n+1)²/4 + 2n(n+1)(2n+1)/6 + n(n+1)/2

= n(n+1)/12 [3n(n+1) + 4(2n+1) +6]

= n(n+1)/12[3n²+3n+8n+4+6]

= n(n+1)/12[3n²+11n+10]

हर का nवां पद = n²(n+1) = n³+n²

हर के n पदों का योग = s₂ =Σn³+ Σn²

= n²(n+1)²/4 + n(n+1)(2n+1)/6

=n(n+1)/12[3n(n+1)+2(2n+1)]

= n(n+1)/12[3n²+3n+4n+2]

= n(n+1)12[3n²+7n+2]

s₁/s₂ = 3n+5/3n+1

Answer 2

दर्शाया की    $\dfrac{1 \times 2^2 + 2\times 3^2 + ... + n(n + 1)^2 }{1^2 \times 2 + 2^2 \times 3 + ... + n^2(n + 1)} = \dfrac{3n + 5}{3n + 1}$

Step-by-step explanation:

$\dfrac{1 \times 2^2 + 2\times 3^2 + ... + n(n + 1)^2 }{1^2 \times 2 + 2^2 \times 3 + ... + n^2(n + 1)} = \dfrac{3n + 5}{3n + 1}$

aₙ  =  n(n + 1)²/(n²(n + 1))

∑ aₙ  = ∑n(n + 1)² /∑ (n²(n + 1))

∑n(n + 1)²  =  ∑ n(n² + 2n + 1)

= ∑ n³ + 2n² + n)

= ∑ n³  + 2∑n² + ∑n

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

= ( n(n+1)/12) ( 3n(n+1)  +  4(2n + 1)  + 6)

= ( n(n+1)/12) ( 3n² + 3n  +  8n + 4 + 6)

= ( n(n+1)/12) ( 3n² + 11n +10)

= ( n(n+1)/12) ( 3n² + 6n  + 5n +10)

= ( n(n+1)/12) (n + 2)(3n + 5)

= n(n + 1)(n+2)(3n + 5)/12

∑ (n²(n + 1) = ∑  (n³ + n²)

= ∑ n³  + ∑n²

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

= ( n(n+1)/12) ( 3n(n+1)  + 2(2n + 1) )

= ( n(n+1)/12) ( 3n²+ 3n  + 4n +2 )

= ( n(n+1)/12) ( 3n²+ 7n +2 )

= ( n(n+1)/12) ( 3n²+ 6n + n +2 )

= ( n(n+1)/12) (n +2 ) (3n + 1)

=  n(n + 1)(n+2)(3n + 1)/12

(n(n + 1)(n+2)(3n + 5)/12)/ n(n + 1)(n+2)(3n + 1)/12

= (3n + 5)/(3n + 1)

सिद्ध किया की

$\dfrac{1 \times 2^2 + 2\times 3^2 + ... + n(n + 1)^2 }{1^2 \times 2 + 2^2 \times 3 + ... + n^2(n + 1)} = \dfrac{3n + 5}{3n + 1}$

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