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 value of the sum of 9+ 16 + 25 + 36 + ….+ 100.

9+ 16 + 25 + 36 + ….+ 100 के जोड़ का मान ज्ञात करिए ?