1.

(A) 1440015. The value of (1° +2° +3 + ... +15') – (1 +2 +3 + ... +15) is(B) 14200(C) 14280(D) 14520​

Answer»

ANSWER:

ANSWER

(1

3

+2

3

+3

3

+.......15

3

)(1+2+3+...15)

SUM of CUBES=

2

n×(n+1)

2

Sum of the AP :

2

n(n+1)

=

2

15(15+1)

2

−

2

15×16

=

2

16×15

2

−15×8

= 15×8

2

−15×8

=14400−120

=14280


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