Question

if
1x2+2x3 + 3x4 +4x5 + ... upto n terms/
1 +2 +3 +4 + ... upto n term=100/3, find n​

Answer 1

Solution :-

→ (1x2+2x3 + 3x4 +4x5 + ... upto n terms)/

(1 +2 +3 +4 + ... upto n term) = 100/3

Solving LHS , Numerator Part first :-

→ 1*2 + 2*3 + 3*4 + 4*5 + __________ upto n terms.

we have :-

• 1st Term = 1*2
• 2nd Term = 2*3
• 3rd Term = 3*4
• nth term = n(n+1) = n² + n

So,

→ sum of n terms = ⅀(n² + n)

→ Sn = ⅀n² + ⅀n

→ Sn = [n(n+1)(2n+1)/6] + [n(n+1)/2]

→ Sn = {n(n+1)/2} [ (2n+1)/3 + 1 ]

→ Sn = {n(n+1)/2} [ (2n + 1 + 3) / 3]

→ Sn = {n(n+1)/2} [(2n + 4)/3]

→ Sn = {2n(n+1)(n+2)} / 6

Now, Denominator Part :-

→ 1 + 2 + 3 + 4 + ________ Upto n terms.

→ Sn = n(n+1)/2

Putting Both values Now, we get,

→ [{2n(n+1)(n+2)} / 6] / {n(n+1)/2} = 100/3

→ [{2n(n+1)(n+2)} / 6] * 2/n(n+1) = 100/3

→ 2(n+2) / 3 = 100/3

→ 2(n + 2) = 100

→ 2n + 4 = 100

→ 2n = 100 - 4

→ 2n = 96

→ n = 48 . (Ans.)

Answer 2

Step-by-step explanation:

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