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Let $P =\sum_{\substack{1\le i \le 2k \\ i\;odd}} i$ and $Q = \sum_{\substack{1 \le i \le 2k \\ i\;even}} i$, where $k$ is a positive integer. Then

1. $P = Q - k$
2. $P = Q + k$
3. $P = Q$
4. $P = Q + 2k$
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$\textbf{P}=1+3+5+7+.....+(2k-1)$

$=(2-1)+(4-1)+(6-1)+(8-1)+.......+(2k-1)$

$=(2+4+6+8+.....2k)+(-1-1-1-1-1.....k \ \text{times})$

$=\textbf{Q}+(-k)=\textbf{Q-k}$
edited by
Substitute k=3 then we get p=9 and q=12  on verifying we get option A.
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That's true. P is adding the first k odd numbers and Q is adding the first k even numbers. Each even number is 1 greater than its corresponding odd number being added. So for k additions, Q will be P+k
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but option c also satisfies given condition........................... checking for each k
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@sanjay how option c will true please explain?
The odd series is 1 3 5 7 ... 2k-1

So, 1+(t1-1)2=2k-1

or t1=k;

P = (k/2) [2x1+(k-1)2]=k^2

The even series is 2 4 6 8 10  ... 2k

So, 2+(t2-1)2=2k

or t2=k;

Q = (k/2) [2x2+(k-1)2]=k^2+k
edited
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Nice approach.Thanks :-)

Just small correction needed."The odd series is 2 4 6 8 10  ... 2k ."