1 k -2 k -3 k - n k is divisible by 1-2-3- n for every n - N- then k is-A- evenB- oddC- only primeD- none of these

The correct option is B oddCheck for different values of k. 1^1 + 2^1 + 3^1 + 4^1/1 + 2 + 3 + 4 = 1 1^3 + 2^3 + 3^3 + 4^3/1+2+3+4 = 100/10 = 10 and so on. You ...

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Quantitative Aptitude

Series

1 k +2 k +3 k...

Question

$1^{k} + 2^{k} + 3^{k} + . . . + n^{k}$ is divisible by 1 + 2 + 3+... n for every $n ϵ N,$ then k is:

even

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odd

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only prime

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none of these

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Solution

The correct option is B odd
Check for different values of k.
$\frac{1^{1} + 2^{1} + 3^{1} + 4^{1}}{1 + 2 + 3 + 4} = 1 \frac{1^{3} + 2^{3} + 3^{3} + 4^{3}}{1 + 2 + 3 + 4} = \frac{100}{10} = 10$
and so on.
You can consider some more examples and then you will find that only choice (b) is correct.

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Similar questions

{lim}_{n \to \infty} \frac{(1^{k} + 2^{k} + 3^{k} + . . . . . + n^{k})}{(1^{2} + 2^{2} + . . . . . + n^{2}) (1^{3} + 2^{3} + . . . . . + n^{3})} = F (k)

, then

(k \in N)

Q. If k is and n be + ive integers and

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

then show that

\sum_{r = 1}^{m}^{m + 1} C_{r} S_{r} =, (n + 1)^{m + 1} - (n + 1) .

Hence evaluate

s_{4} .

Q. If k and n are positive integers and

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

then

\sum_{r = 1}^{m}

^{(m + 1)} C_{r} S_{r}

Q. Solve:

l i m n \to θ (\frac{1^{k} + 2^{k} + 3^{k} + . . . . . k}{n^{k + 1}}) = (k > - 1)

Q. If

P (n) :^{''} 2^{2 n} - 1

is divisible by

k

for all

n \in N^{''}

is true, then the value of

^{'} k^{'}

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1k+2k+3k+...+nk is divisible by 1 + 2 + 3+... n for every nϵN, then k is:

The correct option is B oddCheck for different values of k. 11+21+31+411+2+3+4=113+23+33+431+2+3+4=10010=10 and so on. You can consider some more examples and then you will find that only choice (b) is correct.

$1^{k} + 2^{k} + 3^{k} + . . . + n^{k}$ is divisible by 1 + 2 + 3+... n for every $n ϵ N,$ then k is:

The correct option is B odd
Check for different values of k.
$\frac{1^{1} + 2^{1} + 3^{1} + 4^{1}}{1 + 2 + 3 + 4} = 1 \frac{1^{3} + 2^{3} + 3^{3} + 4^{3}}{1 + 2 + 3 + 4} = \frac{100}{10} = 10$
and so on.
You can consider some more examples and then you will find that only choice (b) is correct.