# Polynomials

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## Summation

6 + 24 + 60 + 120 + 210 + 336 + 504 + 720…. upto 10 terms is equal to?
1. 3680
2. 4290
3. 5720
4. 6170

Choice B. 4290

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## Detailed Solution

= 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + 4 × 5 × 6 + 5 × 6 × 7....
= n (n + 1) (n + 2)
Or, (m – 1) m (m + 1) where m = n + 1
= m3 – m
Sum of all these = ${(\frac{(m(m+1))}{2})^2 - (\frac{(m(m+1))}{2}) = (\frac{(m(m+1))}{2}) * (\frac{(m(m+1))}{2} - 1) }$
= $(\frac{(m(m+1)(m^2+m+1))}{4})$
Substituting m = n + 1, we get $(\frac{(n(n+1)(n+2)(n+3))}{4})$. ….. Surprisingly enough, this seems to work!!!
Substituting n = 10,
We get $(\frac{(10(11)(12)(13))}{4})$.
= 10 × 11 × 3 × 13 = 4290

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