find the sum to n terms of the series
1.2^2 +3.3^2+5.4^2 +... to n terms

Question

find the sum to n terms of the series
1.2^2 +3.3^2+5.4^2 +... to n terms

kc_kennylau · Answer

$$1.2^2+3.3^2+5.4^2+...+(2.1n-0.9)^2$$

kc_kennylau · Answer

(Just for reference, this isn't the answer)

ganeshie8 · Answer

is $.$ a decimal or multiplication symbol ?

anonymous · Answer

i'll finish kc_kennylau's answer:
$$S = 1.2^2 + 3.3^2 + 5.4^2 + .. + (2.1n - 0.9)^2 = \sum_{k=1}^{n} (2.1k - 0.9)^2$$
$$= \sum_{k=1}^{n} 4.41k^2 - \sum_{k=1}^{n}3.78k + \sum_{k=1}^{n}8.81$$
= 4.41 (1/6)n (n+1) (2n+1) - 3.78 (n/2)(n+1) + 8.81n
= (3/200) (n) (98n^2 - 21n + 23)

anonymous · Answer

last term should be 0.81 (not 8.81, typo)

anonymous · Answer

the answer given is (1/2)n(n^3+4n^2+4n-1)  ... which does not match

primeralph · Answer

I'm guessing they confused your period for a decimal point.

primeralph · Answer

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