If _k=1^na_k=6n^3, then _k=1^6(a_k+1-a_k36)^2 is equal to ________.

Question

If $\sum _{k=1}^{n}{a}_{k}=6{n}^{3}$, then $\sum _{k=1}^{6}{(\frac{{a}_{k+1}-{a}_{k}}{36})}^{2}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 91

Accepted Answer

${a}_{1}+{a}_{2}+\cdots +{a}_{n}=6{n}^{3}.$

To get a formula for one term, write the sum up to $n+1$ terms:

${a}_{1}+{a}_{2}+\cdots +{a}_{n}+{a}_{n+1}=6(n+1{)}^{3}$

Now subtract the first equation from this to isolate ${a}_{n+1}$:

${a}_{n+1}=6(n+1{)}^{3}-6{n}^{3}$

Use the identity ${x}^{3}-{y}^{3}=(x-y)({x}^{2}+xy+{y}^{2})$ with $x=n+1$ and $y=n$:

$=6((n+1)-n)((n+1{)}^{2}+{n}^{2}+n(n+1))$

Since $(n+1)-n=1$, simplify the bracket:

${a}_{n+1}=6(1)(3{n}^{2}+3n+1)$

So, replacing $n$ by $n-1$ gives a formula for ${a}_{n}$:

${a}_{n}=6(3(n-1{)}^{2}+3(n-1)+1)$

Now simplify inside:

$=6(3({n}^{2}-2n+1+n-1)+1)$

Combine terms:

$=6(3({n}^{2}-n)+1)$

So we get:

${a}_{n}=6(3{n}^{2}-3n+1)$

Now compute ${a}_{k+1}-{a}_{k}$ using the formulas:

$\sum _{k=1}^{6}{(\frac{6(3{k}^{2}+3k+1)-6(3{k}^{2}-3k+1)}{36})}^{2}$

Inside the bracket, cancel common parts:

$\sum _{k=1}^{6}{(\frac{36k}{36})}^{2}=\sum _{k=1}^{6}{k}^{2}=\frac{6	imes 7	imes 13}{6}=91$

Solutions · 1