IF 1+2+3+......+K IS A PERFECT SQUARE N^2 WHERE N<100 THE POSSIBLE VALUES FOR 'K' IS/ARE..............

Question

IT'S A PRMO SAMPLE QUESTION

Answer 1

a perfect square times a perfect square equals a perfect square

the sum of the integers is k(k+1)/2 and N<100

therefore k cannot be greater than 140 because 140(141)/2=9870 and √9870=99.3478

look at all the perfect squares 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121

you can't go higher than 121 because 144(145)/2=10,440 and √10,440=102.176 and N>100 (102.176 is not a perfect square either)

look at each perfect square listed and see if half of the previous consecutive number or half of the next consecutive number is a perfect square (the sum again is k(k+1)/2)

k can be 8 because 9 is a perfect square and (1/2)8=4 is a perfect square

1+2+3+4+5+6+7+8=36 and 8*9/2=36

k can be 49 because (1/2)50=25 is a perfect square

1+2+3+...+47+48+49=49*50/2=1225 and √1225=35

the values for k are 8 and 49

 

 

Answer 2

S_n = n(n+1) / 2         sum of integers

n² = n(n + 1) / 2

2n² = n² + n

n² = n

 

Only possible value for K is 1.