Question

Find the sum of
1².n + 2².(n-1) + 3²(n-2) + 4²(n-3) +.... to n terms.

Answer 1

Answer:

n(n + 1)²[n + 2]/12

Step-by-step explanation:

Hi,

GIven Sn = 1².n + 2².(n-1) + 3²(n-2) + 4²(n-3) +.... to n terms.

= ∑r².(n - r + 1)

= ∑r²(n + 1 - r)

= (n + 1)∑r² - ∑r³------------(1)

∑r² = Sum of squares of first n natural numbers

= n*(n + 1)*(2n + 1)/6---------(2)

∑r³ = Sum of cubes of first n natural numbers

= n²(n + 1)²/4-----------(3)

Substituting (2) and (3) in (1), we get

Sn = (n + 1)[n(n+1)(2n + 1)/6] - n²(n + 1)²/4

= n(n + 1)²[(2n + 1)/6 - n/4]

= n(n + 1)²[4n + 2 - 3n]/12

= n(n + 1)²[n + 2]/12

Hope, it helps !