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Question : 74 of 160
Marks:
+1,
-0
Solution:
If we consider the option (a), we have
‌‌‌=‌| ak(1k+2k+3k+....+nk) |
| nkâ‹…n |
‌=‌‌⋅‌‌=‌ak(‌)k⋅‌ On putting
‌→dx,‌→x and
Σ→∫=ak⋅xk⋅dx
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