Prove that S2n+1 = (2n+1)an+1
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Solution :
S1S2=2n+1n+1
Sn=an+n(n−1)2⋅d
S1=a+(2n+11)+(2n+1−1)d2=[2n+1][a+nd=−(1)
Number of terms in S2=n+1
S2=(n+1)a+(n+1)(n+1−1)2(2d)=(n+1)a+(n+1)nd
=(n+1)(a+nd)
S1S2=[2n+1][a+nd][n+1][a+nd]=2n+1n+1.
Step-by-step explanation:
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