प्रश्न 1 से 7 तक प्रत्येक श्रेणी के n पदों का योग ज्ञात कीजिए। 
Answers
Answer:
Step-by-step explanation:
यहाँ, दी गयी शृंखला : 1^2 + (1^2 + 2^2 ) + (1^2 + 2^2 + 3^2) + ...
n वां पद = a_n = 1^2 + 2^2 + 3^2 + ...n^2 = [ n (n + 1 ) ( 2n + 1) / 6
= 1/6 [n (2n^2 + 3n + 1 )]
= 1/6 [2n^3 + 3n^2 + n]
= (1/3)n^3 +( 1/2)n^2 + (1/6) n, इसलिये
S_n = ∑_{n=1}^{n} (an) = ∑_{n=1}^{n} [ (1/3)n^3 +( 1/2)n^2 + (1/6) n ]
= 1/3 ∑_{n=1}^{n} (n^3) + 1/2 ∑_{n=1}^{n} (n^2) + 1/6 ∑_{n=1}^{n} (n)
= 1/3 [ n (n + 1 / 2 ]^2 + 1/2 [ (n + 1) (2n + 1) / 6 ] + 1/6 [ n (n+1) / 2]
= n (n+1)/12 [n ( n + 1) + ( 2n +1) + 1 ]
= n (n+1)/ 12 [ n^2 + n + 2n + 1 + 1 ]
= n ( n + 1 ) ( n^2 + 3n + 2 ) / 12
= n ( n + 1 ) ( n + 1 ) ( n + 2 ) / 12
= n ( n + 1 )^2 ( n + 2 ) / 12
इसप्रकार, दी गयी श्रेणी के n पदों का योग n ( n + 1 )^2 ( n + 2 ) / 12 है |
n(n+1)² (n+2)/12 = 1² + (1² + 2²) + (1² + 2² + 3²) ......................
Step-by-step explanation:
1² + (1² + 2²) + (1² + 2² + 3²) ......................
aₙ = (1² + 2² + 3² +..................+ n²)
∑aₙ = (n)(n + 1)(2n + 1)/6
Sₙ = ∑ ∑aₙ 1 से n
Sₙ= ∑ (n)(n + 1)(2n + 1)/6 1 से n
= ∑ (n² + n)(2n + 1)/6
= ∑ (2n³ + 3n² + n)/6
= (1/6) ( ∑ 2n³ + ∑ 3n²+ ∑n ) 1 से n
= (1/6) ( 2∑n³ + 3∑n²+ ∑n )
∑n³ =( (n)(n + 1)/2)²
∑n² = (n)(n + 1)(2n + 1)/6
∑n = n(n+1)/2
Sₙ= (1/6) ( 2( (n)(n + 1)/2)² + 3(n)(n + 1)(2n + 1)/6 + n(n+1)/2)
=> Sₙ= (1/6) ( (n²(n + 1)²/2 + (n)(n + 1)(2n + 1)/2 + n(n+1)/2)
=> Sₙ= (1/12)n(n+1) ( (n(n + 1) + (2n + 1) + 1)
=> Sₙ= (1/12)n(n+1) (n² + n + 2n + 1 + 1)
=> Sₙ= (1/12)n(n+1) (n² + 3n + 2)
=> Sₙ= (1/12)n(n+1) (n+2)(n+1)
=> Sₙ= (1/12)n(n+1)² (n+2)
=> Sₙ= n(n+1)² (n+2)/12
1² + (1² + 2²) + (1² + 2² + 3²) ...................... + = n(n+1)² (n+2)/12
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