find the sum of n termsin the series
1.2+2.3+3.4+.............(n-1)n
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The series is 1.2 + 2.3 + 3.4 + .....+ (n-1)n + n(n+1)
From the given series, the nth term is n(n+1) or (n² + n)
Sum of first n terms of a series with nth term (n² + n) = Sum of first n terms of a series with nth term (n²) + Sum of first n terms of a series with nth term (n)
![= \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ \\ = \frac{n(n+1)(2n+1)+3n(n+1)}{6} \\ \\ = \frac{n(n+1)[(2n+1)+3]}{6} \\ \\ = \frac{n(n+1)(2n+4)}{6} \\ \\ = \frac{n(n+1)(n+2)}{3}](https://tex.z-dn.net/?f=%3D+%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B6%7D+%2B+%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D++%5C%5C++%5C%5C+%3D++%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%2B3n%28n%2B1%29%7D%7B6%7D+%5C%5C++%5C%5C+%3D+%5Cfrac%7Bn%28n%2B1%29%5B%282n%2B1%29%2B3%5D%7D%7B6%7D+%5C%5C++%5C%5C+%3D+%5Cfrac%7Bn%28n%2B1%29%282n%2B4%29%7D%7B6%7D+%5C%5C++%5C%5C+%3D+%5Cfrac%7Bn%28n%2B1%29%28n%2B2%29%7D%7B3%7D "= \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ \\ = \frac{n(n+1)(2n+1)+3n(n+1)}{6} \\ \\ = \frac{n(n+1)[(2n+1)+3]}{6} \\ \\ = \frac{n(n+1)(2n+4)}{6} \\ \\ = \frac{n(n+1)(n+2)}{3}")
From the given series, the nth term is n(n+1) or (n² + n)
Sum of first n terms of a series with nth term (n² + n) = Sum of first n terms of a series with nth term (n²) + Sum of first n terms of a series with nth term (n)
raoatchut191:
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yamini, sir???
yeah sirji
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