Let Tn- n2-1 n- and Sn-T1-T2-T3- -Tn- If T10S10-ab- where a and b are relatively prime natural numbers- then the value of b-a is

T_n= ( n^2+1 )n! = [ ( n+1 )n ( n 1 ) ]n! ⇒ T_n=n (n+1)! ( n 1 )n! T_1=1×2! 0 T_2 = 2×3! 1×2! T_3 = 3× 4! 2× 3! T_4 = 4× 5! 3× 4! · · ...

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Permutations and Combinations

Let Tn= n2+1 ...

Question

Let $T_{n} = (n^{2} + 1) n!$ and $S_{n} = T_{1} + T_{2} + T_{3} + \dots + T_{n} .$ If $\frac{T_{10}}{S_{10}} = \frac{a}{b},$ where $a$ and $b$ are relatively prime natural numbers, then the value of $(b - a)$ is

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Solution

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Similar questions

T_{n} = (n^{2} + 1) n!

S_{n} = T_{1} + T_{2} + T_{3} + \dots + T_{n} .

\frac{T_{10}}{S_{10}} = \frac{a}{b},

a

b

(b - a)

T_{n} = n! (n^{2} + 1)

S_{n} = T_{1} + T_{2} + T_{3} + . . . . . T_{n}

\frac{T_{10}}{S_{10}} = \frac{a}{b}

T_{1} = T_{2} = 1

T_{1}, T_{2}, T_{3}, . . ., T_{n}

T_{n + 2} = (T_{n + 1})^{- 1} + T_{n}

n = 1, 2, 3

T_{2019}

T_{1}, T_{2}, T_{3}, \dots

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

S_{2} = T_{2} + T_{4} + T_{6} + \dots + T_{n - 1}

n

\frac{S_{1}}{S_{2}}

T_{1}, T_{2}, T_{3}, \dots

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

S_{2} = T_{2} + T_{4} + T_{6} + \dots + T_{n - 1}

n

\frac{S_{1}}{S_{2}}

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