Page 65 - Mathematics Class - XI

3. Total number of thermocol balls in enclosure III = 1 + 2 + 3 = 1 + 8 + 27 = 36
3
3
3
4. Total number of thermocol balls in enclosure IV = 1 + 2 + 3 + 4 = 1 + 8 + 27 + 64 = 100
3
3
3
3
5. Total number of thermocol balls in enclosure V = 1 + 2 + 3 + 4 + 5 = 1 + 8 + 27 + 64 + 125 = 225
3
3
3
3
3
OBSERVATION
1. Number of balls in enclosure I = 1 = 1 =    12×  2
3

2 
+
 23×  2  2 × ( 2 1  ) 2
2. Total number of balls in enclosure II = 1 + 2 = 9 =   2   =  2 
3
3


+
3. Total number of balls in enclosure III = 1 + 2 + 3 = 36 =  34×  2 =    3× ( 3 1  )  2
3
3
3



2 

2
 45×  2  4 × ( 4 1+  ) 2
4. Total number of balls in enclosure IV = 1 + 2 + 3 + 4 = 100 =   2  =   2 
3
3
3
3


+
 56×  2  5× ( 5 1  ) 2
5. Total number of balls in enclosure V = 1 + 2 + 3 + 4 + 5 = 225 =   2   =  2 
3
3
3
3
3


CONCLUSION
This activity concludes that the sum of cubes of first n natural numbers,
n +  )1
i.e., 1 + 2 + 3 + ... + n =  n × ( 2   2
3
3
3
3


APPLICATION
This activity can be used to solve problems of summation of series.
Knowledge Booster
In a series a + a + a + ... + a + ...
1
n
2
3
(i) If the differences a – a , a – a , a – a ... are in A.P., then the n term is given by
th
1
3
2
2
4
3
a = an + bn + c, where a, b, c are constants
2
n
(ii) If the differences a – a , a – a , a – a ... are in G.P. with constant ratio r, then
3
1
2
3
4
2
a = ar n – 1 + bn + c, where a, b, c are constants
n
To determine constants a, b, c we put n = 1, 2, 3 and equate them with the values of
corresponding terms of the given series.
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