Question

Find the sum of 12 – 22 + 32 – 42 + 52 - 62 + up to 2n terms​

Answer 1

Answer:

We want to find 1

We want to find 1 2

We want to find 1 2 −2

We want to find 1 2 −2 2

We want to find 1 2 −2 2 +3

We want to find 1 2 −2 2 +3 2

We want to find 1 2 −2 2 +3 2 −4

We want to find 1 2 −2 2 +3 2 −4 2

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n terms

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum =

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2n

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2n

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2n [2a+(n−1)d]

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2n [2a+(n−1)d]=

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2n [2a+(n−1)d]= 2

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2n [2a+(n−1)d]= 2n

We want to find 1 2 −2 2 +3 2 −4 2 +.... to 2n terms=1−4+9−16+25−.... to 2n terms=(1−4)+(9−16)+(25−36)+.... to n terms. (after grouping)=−3+(−7)+(−11)+....n termsNow, the above series is in an A.P. with first term a=−3 and common difference d=−4Therefore, the required sum = 2n [2a+(n−1)d]= 2n