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Question : 3 of 160
Marks:
+1,
-0
Solution:
12+32+52+.....+(2n−1)2 = Sum of squares of n odd natural numbers
= [{12+22+32+42+52+... +(2n−1)2+(2n)2}−{22+42+62+.... +(2n)2}] =
−22[] = − =[4n+1−(2n+2)] = = or(∑(2n)2−22‌∑(n)2=)
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