How to solve this ??
Sathya , 13 Years ago
Grade 10
Discuss with colleagues and IITians> SOLVE...
3 Answers
Pavan Kumar
Last Activity: 13 Years ago
Here tn = (2n–1)(2n+1)(2n+3)
∴ Sn = tn = Σ(2n–1)(2n+1)(2n+3)
= Σ((2n)2 –12) (2n+3)
= Σ(4n2 –1)(2n+3)
= Σ(8n3 + 12 n2 – 2n – 3)
= 8 n2(n+1)2/4 + 12 n(n+1)(2n+1)/6 – 2 n(n+1)/2 – 3n
= n[2n (n+1)2 + 2(n+1)(2n+1)–(n+1) –3]
= n[2n (n2 + 2n + 1) +2(2n2 + 3n + 1) – n – 4]
= n[2n3 + 4n2 + 2n2 + 4n2 + 6n + 2 – n – 4]
= n[2n3 + 8n2 + 7n – 2]
Hence Sn = n[2n3 + 8n2 + 7n – 2]
Sathya
Last Activity: 13 Years ago
Thnks for ur reply...
But if n=1 ,thn the the value of the term should be the value of first term of this expression....
Substituting n=1 in Sn = n[2n3 + 8n2 + 7n – 2] , we get 15....
But the value of first term is 1/1.3=1/3....
Then how can it be the answer?
Pavan Kumar
Last Activity: 13 Years ago
Sorry I read the question wrong here is the solution
tr = 1/r(r+1)(r+2)(r+3)
tr+1 = 1/(r+1)(r+2)(r+3)(r+4)
tr/tr+1 = r+4/r
r tr = (r + 4) tr+1
r tr = (r + 1) tr+1 + 3 tr+1
Putting r = 1, 2, ……, n–1
adding we get, t1 – ntn = 3 [t2 + t2 + t3 +…+ tn]
or 4t1 – ntn = 3[t1 + t2 +…+ tn] = 3 Sn
Sn = 1/3 [4t1 – ntn]
=1/3 [1/1.2.3 – m/3(n+1)(n+2)(n+3)]
= 1/18 – 1/3(n+1)(n+2)(n+3) (Ans.)
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