If Σ_k=1^n a_k = 6n^3, then Σ_k=1^6 ( a_k+1 - a_k 36 )^2 is equal to .

Given $\displaystyle \displaystyle\sum_{k=1}^{n} a_k = 6n^3$k=1∑n​ak​=6n3.

Let $S_n = 6n^3$Sn​=6n3. The $n$n-th term of the sequence is given by $a_n = S_n - S_{n-1}$an​=Sn​−Sn−1​.

$$a_n = 6n^3 - 6(n-1)^3$$an​=6n3−6(n−1)3

$$a_n = 6[n^3 - (n^3 - 3n^2 + 3n - 1)]$$an​=6[n3−(n3−3n2+3n−1)]

$$a_n = 6(3n^2 - 3n + 1)$$an​=6(3n2−3n+1)

Replacing $n$n with $k$k and $k+1$k+1, we get:

$$a_k = 6(3k^2 - 3k + 1)$$ak​=6(3k2−3k+1)

$$a_{k+1} = 6(3(k+1)^2 - 3(k+1) + 1) = 6(3k^2 + 3k + 1)$$ak+1​=6(3(k+1)2−3(k+1)+1)=6(3k2+3k+1)

Now, find the difference $a_{k+1} - a_k$ak+1​−ak​:

$$a_{k+1} - a_k = 6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1)$$ak+1​−ak​=6(3k2+3k+1)−6(3k2−3k+1)

$$a_{k+1} - a_k = 6(6k) = 36k$$ak+1​−ak​=6(6k)=36k

Substitute this into the given summation:

$$\displaystyle\sum_{k=1}^{6} \left(\dfrac{a_{k+1} - a_k}{36}\right)^2 = \sum_{k=1}^{6} \left(\dfrac{36k}{36}\right)^2$$k=1∑6​(36ak+1​−ak​​)2=k=1∑6​(3636k​)2

$$\Rightarrow \displaystyle\sum_{k=1}^{6} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2$$⇒k=1∑6​k2=12+22+32+42+52+62

Using the formula for the sum of squares of the first $n$n natural numbers, $\displaystyle \displaystyle\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$k=1∑n​k2=6n(n+1)(2n+1)​:

$$\displaystyle\sum_{k=1}^{6} k^2 = \dfrac{6(6+1)(2(6)+1)}{6} = \dfrac{6 \times 7 \times 13}{6} = 91$$k=1∑6​k2=66(6+1)(2(6)+1)​=66×7×13​=91

Answer: $91$91