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Answered by
1
Answer:
n²(2n + 5)
Step-by-step explanation:
∑ (6r² + 4r -3)
= ∑ 6r² + ∑4r - ∑3
= 6 n(n + 1)(2n + 1)/6 + 4n(n+1)/2 - 3n
= n(n + 1)(2n + 1) + 2n(n + 1) - 3n
= n(n + 1) (2n + 1 + 2) - 3n
= n(n + 1) (2n + 3) - 3n
= n (n + 1) (2n + 3) - 3)
= n ( 2n² + 5n + 3 - 3)
= n(2n² + 5n)
= n²(2n + 5)
∑ (6r² + 4r -3) = n²(2n + 5)
Answered by
0
Answer:
n²(2n + 5)
Step-by-step explanation:
In this question,
We need to find the
∑ (6r² + 4r - 3)
= ∑ 6r² + ∑4r - ∑3
=
= n(n + 1)(2n + 1) + 2n(n + 1) - 3n
= n(n + 1) (2n + 1 + 2) - 3n
= n(n + 1) (2n + 3) - 3n
= n (n + 1) (2n + 3) - 3)
= n ( 2n² + 5n + 3 - 3)
= n(2n² + 5n)
= n²(2n + 5)
∑ (6r² + 4r -3) = n²(2n + 5)
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