Find the sum of n terms …

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Brainly.in What is your question? Secondary School Math 6 points Find the sum to n terms of series: 5+11+19+29+41.............It is G.P of 11th Std not A.P. Ask for details Follow Report by Rajendrapatel25 27.10.2018 Answers preetam98 preetam98 Ambitious N(n+1)(2n+1/3+1)-1 n(n+1)(2n+4)/6-1 3.6 30 votes THANKS 35 Comments (1) Report Thank you so much Log in to add a comment pr264428 Ambitious Answer: \frac{n(n+1)(2n+4)}{6}-1 Step-by-step explanation: In the given series, We have the series, 5 + 11 + 19 + 29 + 41 + ................. So, We can simply write it as, 5 + 11 + 19 + 29 + 41 + ................. = (2² + 1) + (3² + 2) + (4² + 3) + (5² + 4) + ............ On rearranging, we get, = (2² + 3² + 4² + 5² +.......) + (1 + 2 + 3 + ........) Now, we know that sum of squares of first 'n' terms is given by, Sum=\frac{n(n+1)(2n+1)}{6} So, We can write the first part as, 2^{2} + 3^{2} + 4^{2} + 5^{2} +.......=\frac{n(n+1)(2n+1)}{6}-1 also, Sum of the second half is given by, Sum=\frac{n}{2}(2a+(n-1)d)\\Sum=\frac{n}{2}(2+(n-1))=\frac{n}{2}(n+1) Therefore, the total sum of the 'n' terms is given by, Sum=(\frac{n(n+1)(2n+1)}{6}-1)+\frac{n}{2}(n+1)\\Sum=\frac{n}{2}(n+1)\left[\frac{2n+1}{3}+1\right]-1\\Sum=\frac{n(n+1)(2n+4)}{6}-1 Therefore, the sum of the the terms is given by, Sum=\frac{n(n+1)(2n+4)}{6}-1\\