If sn - C0 C1 - C1 C2 - - Cn -1 Cn and sn - 1sn - 154 then prove that n - 2 - 4

s_n #160; = C_0 C_1 + #160;C_1 C_2 + ... #160; #160;C_n 1 C_n #160; #160; = (2n)!(n 1)!(n + 1)! as prove ...

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If sn = C...

Question

If $s_{n} = C_{0} C_{1} + C_{1} C_{2} + . . . C_{n - 1} C_{n}$ and $\frac{s_{n} + 1}{s_{n}} = \frac{15}{4}$ then prove that n = 2 , 4

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S_{n} =^{n} {C_{0}}^{n} C_{1} +^{n} {C_{1}}^{n} C_{2} + . . . . . +^{n} {C_{n - 1}}^{n} C_{n}

n \in N

\frac{S_{n + 1}}{S_{n}} = \frac{15}{4}

^{'} n^{'}

S_{n} =^{n} C_{0} .^{n} C_{1} +^{n} C_{1} .^{n} C_{2} + . . . . +^{n} C_{n - 1} .^{n} C_{n} a n d \frac{S_{n + 1}}{S_{n}} = \frac{15}{4}

s_{n} = 1 + q + q^{2} + . . . + q^{n}

S_{n} = 1 + \frac{q + 1}{2} + {(\frac{q + 1}{2})}^{2} + . . . + {(\frac{q + 1}{2})}^{n}, q \neq 1

^{n + 1} C_{1} +^{n + 1} C_{2} . s_{1} +^{n + 1} C_{3} . s_{2} + . . . +^{n + 1} C_{n + 1} . s_{n} = k^{n} . S_{n}

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

S_{n} =^{n} C_{0}^{n} C_{1} +^{n} C_{1}^{n} C_{2} + . . . +^{n} C_{n - 1}^{n} C_{n}

\frac{S_{n + 1}}{S_{n}} = \frac{15}{4}

n (n ϵ N)

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