If 1²+2²+3² ……+x² = x(x+1)(2x+1)/6, then 1²+3²+5²…….+19² is equal to ?

यदि 1²+2²+3² ……+x² = x(x+1)(2x+1)/6 हो, तो 1²+3²+5²…….+19² बराबर है ?

Sum of squares of odd numbers from 1 to n = n (n+1)(n+2)/6

As per the question -

n=19

= n (n+1)(n+2)/6

= 19(19+1)(19+2)/6

= 19 x 20 x 21 /6

= 7980/6

= 1330

Hence 1²+3²+5²…….+19² = 1330

Hence the correct answer is option A.

1 से n तक की विषम संख्याओं के वर्गों का योग = n (n+1)(n+2)/6

प्रशानुसार -

n=19

= n (n+1)(n+2)/6

= 19(19+1)(19+2)/6

= 19 x 20 x 21 /6

= 7980/6

= 1330

अतः 1²+3²+5²…….+19² = 1330

अतः सही उत्तर विकल्प A है l

Find the least value of 'x' so that the number '97468x4' is divisible by 8.

'x' का न्यूनतम मान ज्ञात कीजिये जिससे कि संख्या '97468x4', 8 से विभाज्य हो।