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। 

If the sum of two numbers is multiplied by those numbers separately, then the product comes to be 247 and 114 respectively. Accordingly, what is the sum of those numbers?

यदि दो संख्याओं के योग को उन संख्याओं से अलग-अलग गुणा की जाए, तो गुणनफल क्रमश 247 तथा 114 आता है । तदनुसार उन संख्याओं का योगफल कितना है ?