Question

1,3,5,7,...... Find t35 =? And s20=?

Answer 1

Answer:

a+6d=3(a+2d)+2 a+6d=3*7+2 a7 = 21+2 a7 = 23 a+2d=7 a+6d=23 -------------- -4d = -16 d = 4 a = -1. S20 = n/2[2a+(n-1)d] = 10 [2(-1) +19*4]

Step-by-step explanation:

hope it helpful

Answer 2

11∗3∗5+13∗5∗7+15∗7∗9+...  

This can be rewritten as:

∑n=1n1(2n−1)(2n+1)(2n+3)  

Factorizing the term,

14∗∑n=1n4(2n−1)(2n+1)(2n+3)  

⟹14∗∑n=1n(2n+3)−(2n−1)(2n−1)(2n+1)(2n+3)  

⟹14∗∑n=1n[2n+3(2n−1)(2n+1)(2n+3)−2n−1(2n−1)(2n+1)(2n+3)]  

⟹14∗∑n=1n[1(2n−1)(2n+1)−1(2n+1)(2n+3)]  

Simplifying this, we get

14∗[11∗3−13∗5+13∗5−15∗7+...+1(2n−1)(2n+1)−1(2n+1)(2n+3)]  

⟹14∗[11∗3−1(2n+1)(2n+3)]  

⟹(2n+1)(2n+3)−33∗4∗(2n+1)(2n+3)  

⟹4n2+8n+3−33∗4∗(4n2+8n+3)  

⟹4n2+8n3∗4∗(4n2+8n+3)  

⟹n2+2n3(4n2+8n+3)  

⟹n(n+2)3(4n2+8n+3)  

So, the sum of the series is  n(n+2)3(4n2+8n+3)

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