# Find sum of 16 terms ....?

Update:

Could some body tell me - in your country, at what level student are supposed to solve such problems ? Relevance

1^3/1, (1^3 + 2^3)/4, (1^3 + 2^3 + 3^3)/9, (1^3 + 2^3 + 3^3 + 4^3)/16, ...

a_n = 1/4 (n^2 + 2 n + 1)

a16 = 1/4 (289) = 72.25

1 + 9/4 + 4 + 25/4 + ..... + 289/4

sum_(n=1)^16 1/4 (n + 1)^2 = 446

• For me this is a big punishment .Kindly write to me at - pramodkumartandon@yahoo.co.in if possible.

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• High school (age 15 to 18).

Finite sums are a precursor to infinite sums, which are a precursor to calculus.

This problem could be solved simply by typing the whole thing into a calculator (or into a site like wolframalpha). But if we remember some formulas for sums we can come up with an algebraic solution:

*********** FORMULAS ***********

(1 + 3 + 5 + 7 ... + 2k-1) = k^2

n

Σ i^2 = n(n+1)(2n+1) / 6

i=1

n

Σ i^3 = n^2(n+1)^2 / 4

i=1

**************************************

Therefore the terms of the series

an = (1^3 + 2^3 + 3^3 ... + k^3) / (1 + 3 + 5 + 7 ... + 2k-1)

can be written as

an = (k^2 (k+1)^2 / 4) / k^2

= (k+1)^2/4

Now we sum the first 16 terms.

16           17          17

Σ(k+1)^2/4  =  1/4 * Σ k^2   =  1/4 * ((Σ k^2) - 1^2)

k=1         k=2          k=1

= 1/4 * (17(17+1)(2*17+1)/6 - 1)

= 446

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• 1^3/1 + (1^3 + 2^3)/4 + (1^3 + 2^3 + 3^3)/9 + (1^3 + 2^3 + 3^3 + 4^3)/16 + .

T16 = (1^3 + 2^3 + 3^3 + 4^3 .+.+16^3) / [16/2(2+15*2)]

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• in latvia its 7th grade i think.

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• I did not have mathematics in higher secondary level but I read science with anthropology as optional. I appreciate your question and I have respect for a senior citizen like you. For this reason, I gave a reply.

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• 1/1, 9/4, 4, 25/4, ...

a_n = 1/4 (n^2 + 2 n + 1)

a16 = 1/4 (289) = 289/4

sum_(n=1)^16 1/4 (n + 1)^2 = 446

• pramod, you never asked for the steps. You wanted to know at what level are such questions seen.

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• Study of series might be pre-calculus.

My ideas are reasonable but I must have made an arithmetical error somewhere.

Two reliable guys make it 446

Sum of n cubes is known to be [n(n + 1)/2]^2 = n^2* [n + 1)/2]^2

Sum of first n odd numbers is just n^2

The ratio in each term reduces to (1/4)(n + 1)^2

(For example the third term amounts to 16/4 or 36/9 = 4)

4S = 4 + 9 + 16 + 25 + 36 +.. to 16 terms

4S = (1 + 4 + 9 + 16 + 25 + 36 ..to n + 1 terms) - 1

The formula n(n + 1)(2n + 1)/6 is for the sum of the first n squares

With say n = 4 we get  4S = 4*5*9/6 – 1 = 29

but this was only the 4S sum 4 + 9 + 16 which was only 3 terms.

We removed a term when we removed 1 so we need to replace n by n + 1

4S(n) = (n + 2)(n + 3)(2n + 3)/6 – 1

The 4 term sum is calculated as 5*6*11/6 – 1 = 54 = 4 + 9 + 16 + 25

4S(16) = 18(19)(35)/6  - 1 = 1994

So the sum to 16 terms is 498.5

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• This can easily be done at age 12; it involves only multiplications, addition & divisions. Such operations are learned even before age 12. Yes, 16 terms to evaluate would be long to do, but easy.  Now, to find the general formula, or patterns or series representation of formulas, this would be what a 16 or 17 yr old (end highschool or beginning college, calculus class) might get acquainted with.

Note that if u analyze the tops & bottoms, we quickly see the pattern and the sum becomes easy & quick to evaluate.

• Kindly refer to my write up at <<< https://answers.yahoo.com/question/index?qid=20200531021516AAA9A6A>>>

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• That looks like something from a math analysis or maybe precalculus student might see.  Those are different versions of the same level, between 2nd year algebra and a first course in calculus.  I think it's more likely to be found in the "math analysis" version, which is oriented more to eventual math/physical science/engineering majors and is most often taken as a high school junior, with a fair number of sophomores.

This is in the US, in California, as of several years ago.  I don't know how much the "common core" curriculum madness has changed that, though.

Of course, the student is expected to recognize those sums and know that the sum of the first n odd numbers is n² and that ∑k³ = (∑k)².  The nth term is:

a_n = (1² + 2³ + ... + n³) / n² = (∑n)² / n² = [n(n+1)/2]² / n² = (n+1)² / 4

Summing a_n from n=1 to 16 then requires knowing ∑n² = n(n+1)(2n+1)/6.  (...or a lot of button-pushing on a fraction-aware calculator! :^)

• Kindly refer to my write up at <<< https://answers.yahoo.com/question/index?qid=20200531021516AAA9A6A>>>

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