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Brainly.inWhat is your question?Secondary School Math 6 pointsFind 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 27.10.2018Answerspreetam98preetam98 AmbitiousN(n+1)(2n+1/3+1)-1n(n+1)(2n+4)/6-13.630 votesTHANKS 35Comments (1) ReportThank you so much Log in to add a commentpr264428 AmbitiousAnswer:\\frac{n(n+1)(2n+4)}{6}-1Step-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}-1also,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}-1Therefore, the sum of the the terms is given by,Sum=\\frac{n(n+1)(2n+4)}{6}-1\\\\