268 INTEGRATION 12 + 22 + 32 +.. + n2 lim 32. 3 12 n → +% 5 k n-1 2k2 33. lim 34. lim ,2 12 +* k3D1 3 n- +* k=1 Σ (:-) 35. lim n - +x n. in k=1 36. Show that 1.2+2.3+ .+ n(n + 1) = žn(n + %3D 37. Show...

#35 please
268<br>INTEGRATION<br>12 + 22 + 32 +.. + n2<br>lim<br>32.<br>3<br>12<br>n → +%<br>5 k<br>n-1<br>2k2<br>33.<br>lim<br>34.<br>lim<br>,2<br>12 +* k3D1<br>3<br>n- +* k=1<br>Σ<br>(:-)<br>35.<br>lim<br>n - +x<br>n.<br>in<br>k=1<br>36. Show that<br>1.2+2.3+ .+ n(n + 1) = žn(n +<br>%3D<br>37. Show that the sum of the first n consecutive p-<br>integers is n2.<br>When each term of a sum cancels part of the n<br>leaving only portions of the first and last terms at th<br>sum is said to telescope. In Exercises 38-43, eva<br>telescoping sum.<br>V.<br>50<br>38.<br>39.<br>(3k - 3k-<br>k<br>k = 1<br>k + 1<br>k7 |<br>20 (1<br>100<br>

Extracted text: 268 INTEGRATION 12 + 22 + 32 +.. + n2 lim 32. 3 12 n → +% 5 k n-1 2k2 33. lim 34. lim ,2 12 +* k3D1 3 n- +* k=1 Σ (:-) 35. lim n - +x n. in k=1 36. Show that 1.2+2.3+ .+ n(n + 1) = žn(n + %3D 37. Show that the sum of the first n consecutive p- integers is n2. When each term of a sum cancels part of the n leaving only portions of the first and last terms at th sum is said to telescope. In Exercises 38-43, eva telescoping sum. V. 50 38. 39. (3k - 3k- k k = 1 k + 1 k7 | 20 (1 100

Jun 05, 2022
SOLUTION.PDF

Get Answer To This Question

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here