Question

(1) Calculate ?_{i=1}^{2} ?_{j=1}^{3} i^2 j. (2) Calculate ?_{i=5}^{11} i^2 using the table below. TABLE 2 Some Useful Summation Formulae. Sum Closed Form ?_{k=0}^{n} ar^k (r ? 0) ?_{k=1}^{n} k ?_{k=1}^{n} k^2 ?_{k=1}^{n} k^3 ?_{k=0}^{?} x^k, |x| < 1 ?_{k=1}^{?} kx^{k-1}, |x| < 1 [ar^{n+1} - a] / [r - 1], r ? 1 [n(n + 1)] / 2 [n(n + 1)(2n + 1)] / 6 [n^2(n + 1)^2] / 4 1 / (1 - x) 1 / (1 - x)^2

          (1) Calculate ?_{i=1}^{2} ?_{j=1}^{3} i^2 j.
(2) Calculate ?_{i=5}^{11} i^2 using the table below.
TABLE 2 Some Useful Summation Formulae.
Sum
Closed Form
?_{k=0}^{n} ar^k (r ? 0)
?_{k=1}^{n} k
?_{k=1}^{n} k^2
?_{k=1}^{n} k^3
?_{k=0}^{?} x^k, |x| < 1
?_{k=1}^{?} kx^{k-1}, |x| < 1
[ar^{n+1} - a] / [r - 1], r ? 1
[n(n + 1)] / 2
[n(n + 1)(2n + 1)] / 6
[n^2(n + 1)^2] / 4
1 / (1 - x)
1 / (1 - x)^2

(1) Calculate ?i=1^2 ?j=1^3 i^2 j.
(2) Calculate ?i=5^11 i^2 using the table below.
TABLE 2 Some Useful Summation Formulae.
Sum
Closed Form
?k=0^n ar^k (r ? 0)
?k=1^n k
?k=1^n k^2
?k=1^n k^3
?k=0^? x^k, |x| < 1
?k=1^? kx^k-1, |x| < 1
[ar^n+1 - a] / [r - 1], r ? 1
[n(n + 1)] / 2
[n(n + 1)(2n + 1)] / 6
[n^2(n + 1)^2] / 4
1 / (1 - x)
1 / (1 - x)^2

Added by Ramon P.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/02/2023

Step 1

It is likely that the "#" symbol represents a summation, so we can rewrite the expression as: 6 = 1 Σ Ci-1 #j This means we are summing up the values of Ci-1 for all values of j from 1 to 6. Show more…

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