If an-k-0nlog 1 10n-k -n-k - for n - 0 then a0-a1-a2-a3- upto - is equal to t5where - -t- is a least possible two digit number and -s- an even number- Which of the following is correct-

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Standard XII

Mathematics

Combination with Restrictions

If an=∑k=0nlo...

Question

If $a_{n} = \sum_{k = 0}^{n} \frac{(l o g_{e} 10)^{n}}{k! (n - k)!}$ for $n \geq 0$ then $a_{0} + a_{1} + a_{2} + a_{3} + \dots \dots$ upto $\infty$ is equal to $t^{s}$
(where ‘t’ is a least possible two digit number and ‘s’ an even number). Which of the following is correct?

‘t+s’ is an even integer

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‘t+s’ is an odd integer

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‘t+s’ is divisible by 3

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‘t+s’ is divisible by 3

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Solution

The correct options are
A ‘t+s’ is an even integer
C ‘t+s’ is divisible by 3
D ‘t+s’ is divisible by 3
$a_{n} = \frac{(l o g_{e} 10)^{n}}{n!} \sum_{k = 0}^{n} \frac{n!}{k! (n - k)!} = \frac{(l o g_{e} 10)^{n}}{n!} [2^{n}] = \frac{(2 l o g_{e} 10)^{n}}{n!}$
Thus, $a_{0} + a_{1} + a_{2} + \dots \dots$ upto infinity is
$= \sum_{n = 0}^{\infty} \frac{(2 l o g_{e} 10)}{n!} = e^{2 l o g_{e} 10} = 100$

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If an=∑nk=0(loge10)nk!(n−k)! for n≥0 then a0+a1+a2+a3+…… upto ∞ is equal to ts (where ‘t’ is a least possible two digit number and ‘s’ an even number). Which of the following is correct?

The correct options are A ‘t+s’ is an even integer C ‘t+s’ is divisible by 3 D ‘t+s’ is divisible by 3an=(loge10)nn!∑nk=0n!k!(n−k)!=(loge10)nn![2n]=(2 loge10)nn! Thus, a0+a1+a2+…… upto infinity is =∑∞n=0(2 loge10)n!=e2loge10=100

If $a_{n} = \sum_{k = 0}^{n} \frac{(l o g_{e} 10)^{n}}{k! (n - k)!}$ for $n \geq 0$ then $a_{0} + a_{1} + a_{2} + a_{3} + \dots \dots$ upto $\infty$ is equal to $t^{s}$
(where ‘t’ is a least possible two digit number and ‘s’ an even number). Which of the following is correct?