Question 2 Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Class X1 - Maths -Sequences and Series Page 196
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hello users.....
we have to find
the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
solution :-
please see the attachment for answer
⭐⭐ hope it helps ⭐⭐
we have to find
the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
solution :-
please see the attachment for answer
⭐⭐ hope it helps ⭐⭐
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Answered by
1
Let S = 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 .......
we observed nth term = {nth term of 1,2,3...}{nth term of 2,3,4,....}{nth term of 3,4,5,....}
= n(n+1)(n+2)
Tn = n(n+1)(n+2)
= n³ + 3n² + 2n
Now,
Sn = ∑Tn
= ∑( n³ + 3n² + n)
= ∑n³ + 3∑n² + ∑n
= {n(n+1)/2}² + 3n(n+1)(2n+1)/6 + n(n+1)/2
= n(n+1)/2[ n(n+1)/2 + (2n+1) + 1 ]
= n(n+1)/4[(n² + n + 4n + 2 + 2]
= n(n+1)(n² + 5n + 4)/4
we observed nth term = {nth term of 1,2,3...}{nth term of 2,3,4,....}{nth term of 3,4,5,....}
= n(n+1)(n+2)
Tn = n(n+1)(n+2)
= n³ + 3n² + 2n
Now,
Sn = ∑Tn
= ∑( n³ + 3n² + n)
= ∑n³ + 3∑n² + ∑n
= {n(n+1)/2}² + 3n(n+1)(2n+1)/6 + n(n+1)/2
= n(n+1)/2[ n(n+1)/2 + (2n+1) + 1 ]
= n(n+1)/4[(n² + n + 4n + 2 + 2]
= n(n+1)(n² + 5n + 4)/4
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