1.4k views

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. $P = Q + 2k$

retagged | 1.4k views

$\textbf{P}=1+3+5+7+\ldots +(2k-1)$
$\quad=(2-1)+(4-1)+(6-1)+(8-1)+\ldots+(2k-1)$
$\quad=(2+4+6+8+\ldots+2k)+(-1-1-1-1-1\ldots k \ \text{times})$
$\quad=\textbf{Q}+(-k)=\textbf{Q-k}$

Correct Answer: $A$

edited
Substitute k=3 then we get p=9 and q=12  on verifying we get option A.
by
+11
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
0
but option c also satisfies given condition........................... checking for each k
0
@sanjay how option c will true please explain?
0
Lets Assume the value of k=5

then we take number from 1 to 10.

then P= 1+3+5+7+9 = 25

and Q= 2+4+6+8+10 = 30            So, here we conclude that P=Q+K
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
0

Nice approach.Thanks :-)

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

Assme k=3 so 2k=6

p-sum of all odd numbers from 1 to 2k so p=1+3+5=9

q-sum of all even numbers from 1 to 2k so q=2+4+6=12

now as we can see P=Q-K