(A) 14400
15. The value of (1° +2° +3 + ... +15') – (1 +2 +3 + ... +15) is
(B) 14200
(C) 14280
(D) 14520
Answers
Answered by
10
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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